nm0765: er so where we were last time is we had er written down a likelihood 
that we wanted to maximize and then we'd written down the first order condition 
so let's write remember the likelihood the likelihood went like er the 
Lagrangian sorry went like this that was the flow element of it and er there we 
had this extra stuff that came from the end conditions now we realized that 
this Lagrangian didn't contain terms in the X-prime term so this was a 
separable maximization and all we needed to do to maximize this lagra-, 
Lagrangian was to differentiate it at each time into T and set those 
derivatives equal to zero and if we did that what we got from in here was D-F- 
by-D-X plus lambda-D-G- by-D-X is equal to minus-lambda-prime at all times 
little-T okay and that's just differentiating this expression with respect to X 
so that it's maximizing this term in the integral at each time 
little-T with respect to X and the other condition we had was D-F-by-D-U plus 
lambda-D-G- by-D-U was equal to zero at all times little-T and that was again 
differentiating this point thing in the integral here with respect to U at all 
times little-T so we're maximizing this with respect to U now these two 
conditions have a special name they're called the Hamiltonian conditions and er 
the reason they're called the Hamiltonian conditions a-, are apart from being 
named after a man called Hamilton is that they er sta-, are basically 
derivatives of the function F plus lambda times G with respect to X here and 
with respect to U here so we think of the function H which is F plus lambda 
times G as the thing that's being differentiated on the left-hand side now so 
these are like first order conditions if we wanted to maximize something we'd 
take these first order conditions 
and solve them but there's other things here that we also need to worry about 
the other things here that we need to worry about come from differentiating 
this Lagrangian at the terminal time you see X at time T appears here then it 
appears here so if we differentiate the Lagrangian with respect to X at time T 
we don't get an expression like this or an expression like this we get mu minus 
lambda at T and that's actually written right at the bottom of of page two of 
your notes now and this has got to equal zero now where did this bit of the 
Lagrangian come from this bit of the Lagrangian came from the terminal 
constraint by our state variable we had a ke-, terminal constraint which said 
that X at time T had to be greater than or equal to some number X-nought and 
this lambda at time T and this mu at time T are two Lagrange multipliers for X 
at time T mu is the Lagrange multiplier that 
applies this constraint lambda at T is the costate variable so this condition 
here gives you a link between the two so i think probably the best thing to do 
now is to use these conditions to solve a particular problem so let's forget 
this stuff the stuff up there and use the first order conditions to solve a 
particular problem so you can see how they work in practice so what i'm going 
to do is i'm going to take this stuff and put it up there and then i'm going to 
solve the problem over here and you'll be able to see how they relate one to 
the other so first of all i've got to be able to rub the blackboard off excuse 
me right so let's get rid of that and get rid of that and we can get rid of 
that too we'll leave the rest up now the example i give you is on page three of 
the notes and the example is one of a a consumer trying to maximize their 
lifetime's utility so what's the problem the the problem we face is for a 
consumer who lives from time period zero up to time period capital titl-, 
capital-T whose utility is the logarithm of their consumption level at time T 
and er they want to maximize this by choosing a path of consumption and their 
their ability to do this is constrained in certain ways well they start off 
their lives with er a level of saving say er ooh now they've got to have some 
positive saving so let's call it some level S-zero and er they also have a 
constraint on their the way in which saving changes through time so saving 
changes through time that's we look at the derivative of S at T saving's 
going to go up if we get income from our current stock of saving so this is the 
rate of interest on our current stock of saving and saving's going to go down 
if we eat some of our saving right so we're going to maximize our lifetime's 
utility given we start off with this fixed level of saving given saving evolves 
in this way and the final condition we have to have is we have to have we die 
not owning money right right so this is a maximization problem how are we going 
to apply these conditions to this maximization problem well what we want to do 
is we want to think about what is a function F what is our function G here so 
let's do that first of all F at time T er it's an F let's just call it F is a 
logarithm of your consumptions at time T G 
in this particular case it's the stuff here whatever that is in this case it's 
R of S-T minus C of T and that's it so these this is the thing you want to 
maximize this is the thing on the other side of the savings equation this is 
how savings change through time G was the constraint on the rate of change in 
savings clear clear 
sf0766: mm 
nm0765: jolly good so er how are you going to do this well let's do it in two 
stages first of all let's write down a Hamiltonian for this problem so it's the 
logarithm of C plus lambda times R of S minus C that's the er G bit that's the 
F bit now these conditions tell us how to solve the problem now first 
of all we have to do D-H- by- D-X now at where X is the state variable okay so 
let's pick someone namex what's the state variable in this problem what's the 
okay what's the thing we can't choose 
sf0767: consumption [laughter] saving 
nm0765: really 
sf0767: and stuff 
nm0765: really consumption's the thing we're choosing for C for control 
sf0767: it's probably saving 
nm0765: okay S for state very good so the thing we'd we can't control the state 
variable is S so in this case the state variable we call saving so you're going 
to have to do D-H-by-D-S is equal to minus- lambda-prime this one says D-H by 
the control variable which is consumption is 
equal to zero and so we need to take these two equations and now i can rub that 
stuff off and come over here and fill these in in more detail what is H 
differentiated with respect to S well S isn't here but it's it's just this this 
is just the C S only appears here in the H equation so D-H- by-D-S is actually 
lambda times C R sorry and that's got to equal minus-lambda-prime D-H-by-D-C i 
know you like differentiating logarithms so you've got to differentiate this 
logarithm with respect to C one-over-C and there's a C term here so we can have 
minus-lambda and that equals 
zero yes thanks so we've got these two equations that come from the Hamiltonian 
condition but we also know one other thing the other thing we know is how 
saving evolves through time so let's not lose that S-prime of T is equal to R-S 
minus C we've got three equations now i'm going to rub this off and let's see 
what we've got well the first thing we have is that we have a differential 
equation for the costate variable costate variable is comes straight from that 
the second thing we have is we have a relationship between a costate variable 
and consumption at time T and the final thing we have is we have a relationship 
between saving a differential equation for saving and consumption okay so once 
we've 
solved these three equations for the path of the costate variable the path of 
consumption and the path of saving then we will have found the optimal path for 
consumption in this problem and the consequent optimal path for saving okay 
there are some constraints we haven't worried about yet where saving starts up 
where saving finishes we'll bing the bung those in later clear 
sf0768: i don't understand how the Hamiltonian relates to the Lagrangian 
nm0765: [sigh] okay let's go back 
sf0768: sorry 
sf0768: we had the Lagrangian yes 
sf0768: mm 
nm0765: which consisted of F plus what oh dear hang on lambda-G plus some other 
stuff 
sf0768: right 
nm0765: okay and we said we needed to maximize that at each time period little-
T 
sf0768: oh so 
nm0765: so when we maximize the Lagrangian at each pime time period little-T 
we're basically maximizing the ham-, 
Hamiltonian plus some other stuff at each time period little-T 
sf0768: mm 
nm0765: so that's how the Hamiltonian comes in it's just basically a way of 
writing a part of the Lagrangian 
sf0769: er okay 
nm0765: is that all right 
sf0768: so it's not the differential of the Lagrangian then like 
nm0765: no the derivative of the Hamiltonian is the differential of the 
Lagrangian 
sf0768: oh okay 
nm0765: so we differe-, when we differentiate the Lagrangian we're doing the 
same thing we're differentiating the Hamiltonian 
sf0768: so we just forget about all those other bits then Lagrangian 
nm0765: basically provided you know the differential equation that gives you 
this the differential equation that gives you that you can forget all that 
stuff about the Lagrangian if you want if you just want to solve the problem 
sf0768: okay 
nm0765: so the only conditions you need to know i've written on for you at the 
bottom of page two of the notes provided you know that you 
can solve optimal control problems 
sf0768: okey-dokey 
nm0765: okey-dokey now any more questions while i'm at it oh i've got a 
question for you what's the solution for this differential equation i'm going 
to choose the back row ha ha ha you haven't got the faintest idea 
sm0770: solution lambda 
nm0765: ah nice try suppose i gave you X-prime is equal to R-times-X it's a 
linear differential equation with R okay a solution to this one is lambda at 
time T is equal to an arbitrary constant times E-to-the-power-R-T if i 
differentiate this with respect to R i just get an extra R at the front ah 
minus-R is that clear [laugh] i'll take it slower [laughter] i've got to rub 
off the problem 'cause we need all that stuff so we now have these three 
equations in these three unknowns lambda-T we don't know consumption we don't 
know savings 
we don't know we've got differential equation here in lambda it's a very simple 
differential equation it's a differential equation that appeared on right on 
the front of your notes on differential equations it's a one variable linear 
differential equation if you just took those notes on differential equations 
and plugged everything in you would find that the solution was lambda at time T 
some arbitrary constant times E-to-the-minus-R-T you can check that because if 
we do differentiate lambda at time T you get minus-R-K E-to-the-minus-R-T right 
which is the same thing as minus-R times lambda-T yeah brilliant so the 
solution to this equation now i can write it down is lambda at time T is equal 
to an arbitrary constant times E-to-the-minus-R-T so we know 
what lambda is and because we know what lambda is we know what consumption is 
let's get rid of that consumption is one-over- lambda at time T from this 
equation and one-over-lambda at time T is er one-over-K E-to-the-R-T so what's 
happening to the optimal path of consumption in this problem well it's growing 
exponentially fast at the rate E-to-the-R-T we've solved we've learned 
something we don't know where what this arbitrary constant is but we know the 
general pattern of the path of consumption what about savings well the cha-, 
rate in change of savings is equal to R-S minus consumption which is minus- er -
one-over-K E-to-the-R-T so now we have a differential equation for savings at 
time T which we need to solve so do i solve it i've just told you what the 
solution is how are we going to do it well we're 
going to do it in two stages we'll do it S-prime minus R-S is equal to minus-
one-over-K E-to-the-R-T and then we multiply through by E-to-the-minus-R-T so 
we get S-prime E-to-the-minus-R-T minus R E-to-the-minus-R-T S is equal to 
minus-one-over-K and then we notice that this term here is the derivative of S 
times E-to-the-minus-R-T just applying the prod-, product rule so if i 
differentiate S with respect to T and hold the E-to-the-minus-R-T term constant 
i get this bit if i di-, differenti-, hold the S constant and differentiate E-
to-the-minus-R-T with respect to T i get minus-R E-to-the-minus-R-T here and 
then the S which i'm holding constant right just from the product rule do you 
believe me 
sf0771: yeah [laughter] 
nm0765: do you un-, that's another que-, do you understand 
sf0771: yes i think 
nm0765: [laugh] i can do it again 
sf0771: yeah just do the last two lines 
nm0765: okay from this bit to this bit 
sf0771: yeah 
nm0765: okay i said to you the 
problem differentiate this with respect to T what are you going to do well 
first of all you'll differentiate S with respect to T and hold this bit 
constant if you did that you'd get S-prime E-to-the-minus-R-T then you'd hold 
the S bit constant and you'd differentiate this bit so you'd get minus-R E-to-
the-minus-R-T from differentiating this stuff and then the S that you're 
holding constant brilliant 
sf0772: yeah 
nm0765: right so we know the differe-, differential of this with respect to T 
is this so to solve for S oh er you're not going to like that i'd better come 
back over here so for S we say write S E-to-the-minus-R-T is equal to T-over-K 
plus another arbitrary constant which we'll call C or minus-T-over-K just 
integrating both sides we integrate this we get minus-one-over-K times T plus 
another arbitrary constant so finally 
we know the path of saving which is E-to-the-minus-R-T times T-over-K with a 
minus there plus another arbitrary constant er E-to-the-minus-R-T no we don't 
that's a minus-R-T so these are pluses okay so what have we done we've taken 
the differential equation for lambda and solved it that allowed us to find what 
consumption was and then we plugged in the what we knew for consumption into 
the differential equation for saving and solved that so to sum up we've got 
lambda at time T is from here an arbitrary constant times E-to-the-minus-R-T s-,
consumption at time T is the same arbitrary constant one-over-it times E-to-
the-plus-R-T so consumption is growing at the rate exponentially at the rate T 
and we've got saving at time T 
finally from up here which is doing something funny it's growing at the rate E-
to-the-R-T from this bit then we have to knock off K oh sorry knock off T-over-
K times E-to-the-R-T here so it's not clear what saving's doing but if 
consumption's growing i guess saving's got to grow so we've solved our problem 
we took the problem of finding the optimal path for capit-, good consumption 
wrote down the Hamiltonian conditions which gave us this equation and this 
equation and we got er this equation from the constraint solved them and we had 
to solve some differential equations to do that but it's not surprising you 
have to solve differential equations 'cause differential equations are 
equations which have functions as solutions paths as solutions and we're trying 
to find paths aren't we we're trying to find a path for consumption and this is 
what we get right any questions 
sf0768: 
yeah i lost the second when you went from that side of the board over here 
nm0765: this side 
sf0768: yeah 
nm0765: to this side 
sf0768: mm what happened there 
nm0765: what happened there i integrated both sides 
sf0768: oh okay 
nm0765: so if the derivative of this is a constant if anything has a constant 
derivative 
sf0768: mm 
nm0765: then the actual thing has got to be a linear function 
sf0768: right 
nm0765: so this is a has a k-, constant derivative so when i integrate it up 
it's got to have K it's got to be linear in T 
sf0768: okay 
sf0772: is that C consumption or a constant 
nm0765: if i it's a constant 
sf0768: mm okay 
nm0765: so if i do it backwards let's do it backwards if i took that and then i 
differentiated both sides with respect to T what am i going to get if i 
differentiate this with respect to T 
sf0768: 
nm0765: i'm going to get minus-one-over-K which i've 
got here minus-one-over-K here i've got just that differentiated with respect 
to T 
sf0768: mm-hmm 
nm0765: so you're happy that that's the same as that it sorry this stuff is the 
same as this stuff 
sf0768: mm-hmm 
nm0765: so let's just rub out the the brackets oh i'd better write it out again 
S E-to-the-minus-R-T minus-one-over- K T plus an arbitrary constant it's a big 
C not a little C in the notes it's it's clear that it's a big C right but we 
haven't done anything because this solution that we've written up here oh dear 
right this solution that we've written up here has these arbitrary constants 
and how and we got two arbitrary constants we got this K and now we've got this 
C as well this big C here how are we going to 
solve them well we know two things we know what the initial level of saving was 
it's some number S-nought that's what you start your life with you know 
something else as well you know that you can't end your life with zero saving 
so we know that S at time capital-T has got to be greater than or equal to 
nought that gives us a different a condition a la-, Lagrangian which says that 
lambda at time T times S at time T minus zero has got to be greater than or 
equal to zero oh sorry it's got to equal zero we know that S this has got to be 
positive we know that's got to be positive so we can use this condition and 
this condition to tie down the arbitrary constants basically 
what we're going to get is going to get S at time zero is equal to S-nought S 
at time T well right so before we go on any questions yeah 
sm0773: where was that er multiplier you put down on top of the 
nm0765: this one 
sm0773: er that's right 
nm0765: that came from the f-, last things i wrote down right before i rubbed 
off the er Lagrangian the Lagrangian on page two of your notes if you look 
right down at the bottom you'll see what i wrote on the board the very last 
line mu is equal to lambda at time T nu times ek-, ma-, times X at time T-minus-
X-nought oh that's a is equal to zero so this statement here was the last thing 
i wrote down before is equivalent to mu times X at time T- minus-X-nought 
equals zero it's equivalent to that thing okay we're going to deal with these 
terminal 
conditions in greater depth now so let's move on 
nm0765: we've always talked about the constraint at the end as being you can't 
die owe-, owning money you can't die with negative savings but we talked about 
other sorts of terminal constraints that you might have when you're doing these 
optimizations we talked about steering a rocket to the moon and the terminal 
condition there was you ended up at the moon another sort of terminal condition 
you might have is you comes up when you er fill the bath when you fill the bath 
you have a an absolute level that you want the bath to fill to so you know the 
bath is full once the state variable has hit this particular level final sort 
of terminal condition you might come across would occur if you owned a machine 
or a factory or something like that actually this isn't the final one there's 
another one and the terminal condition there is when do you quit when do you 
decide enough is enough and you 
want to leave the factory so that will be a terminal condition req-, required 
you to decide when to stop doing something another ker-, terminal condition 
would arise if once you die or once your machine is dead [laughter] 
om0774: it's okay 
nm0765: [laughter] i'm ruining your equipment [laughter] once your machine is 
dead er you get a certain value from it so er suppose we looked at an 
investment problem and you had to plan your production for your firm so you had 
to choose the path of production and then decide when to scrap your machine but 
once the machine and your investment and your plant were scrapped you could 
take it and sell it to somebody so the amount which you could sell this stuff 
to at the end of the machine's life would determine when you wanted to stop 
okay so there are lots and lots of different sorts of ways in which the end 
point might affect what you wanted to do it 
with your problem so let's look at some of them now er so we're actually on 
page four of your notes so first of all let's study the situation where you 
have a finite time of shall we say death capital-T and what sorts of 
constraints might you face when you die well the easiest would be none what 
that means is that when you die there is no constraint on the your final 
position your final value of the state variable so none I-E X at time capital-T 
is unconstrained if X at time capital-T is unconstrained what would you expect 
the Lagrange multiplier associated with X at time T capital-T to b-, take well 
we'd expect a Lagrange multiplier was zero i-, if if something is unconstrained 
it 
has a zero Lagrange multiplier and that's exactly the condition if you have a 
problem where you don't care about w-, where you end up then it must be the 
case that your costate variable takes a zero value at the end of your life 
second sort of constraint you might have you might have a a precise constraint 
like the filling of the bath which says that once the bath is full it has to 
have a certain amount of water in it and that's it there is no discussion in 
that case lambda at time capital-T is completely unrestricted and we saw that 
sort of constraint coming in when we did er standard constraint optimization 
when we did standard constraint optimizations equality constraints meant lambda-
T could take positive values negative 
values we couldn't necessarily say that Lagrange multiplier will be positive or 
negative when we had equality constraints and the same comes through here final 
condition which we've seen before X at time T has got to be greater than some 
number so that means you can't die owe-, owning money for example and the 
constraint there i've given you is that er that looks wrong to me no lambda at 
time T times X at time T minus X-T equals zero which it so that says either the 
costate variable is zero so this constraint isn't binding and it's like X at 
time T is entirely free or X at time T equals X-subscript- capital-T so the 
constraint does bind and then lambda at time T can be any value so this is a a 
way of writing this statement and this statement in some sort of composite 
form okay so that's w-, these conditions have a prop-, a name they're called 
transversality conditions transversality conditions are conditions that arise 
from the what goes on at the end of a life problem end of the life of a problem 
condition and they're very useful information not in solving for the path 
usually but they're useful information in solving for these arbitrary constants 
nm0765: right now that's er transversality conditions when we have a finite 
time horizon but lots of economic problems we don't want to have finite time 
horizons we want to allow consumers to live forever firms to live forever we 
want to allow countries and planning problems to go on forever and in problems 
like that we need a different set of transversality conditions so let's look at 
er 
infinite horizon problems now infinite horizon problems we really want to say 
the same things as these but for the case where the capital-T is infinite so if 
there's no constraint what we want is then to the limit as T tends to infinity 
of the costate variable of time capital-T is zero so instead of it being zero 
at time t-, capital-T we have it going to zero as we go off to infinity if we 
wanted a constraint which said that in the limit X at time T equals some number 
let's call it X-infinity so as T wen-, goes to infinity the state variable gets 
closer and closer to this number X-infinity then the equivalent transversality 
condition this one over here is just er well let's see 
lim- T tends to infinity lambda at time T is unrestricted and the final one 
again just rewriting that do you think for T tending to infinity is the lim-T 
tends to infinity X at time T is greater than or equal to some number X let's 
call it X-subscript-infinity what we require is that the limit as the capital-T 
tends to infinity lambda-T times X-T minus X-infinity equals zero and all 
that's written in your notes and all this is the same as all that just letting 
capital-T go off places right so we've dealt with three sorts of end point 
restrictions that you might actually have in the your problems and these three 
sorts of end point restrictions just come from the Lagrange multiplier nature 
of the problem and we've done them in two cases we've done them where we in 
finite time horizons we've done them for infinite time horizons but there are 
other sorts of end point constraints that 
we talked about so let's look at one of them now now let's think of the machine 
problem the machine problem what you have is you have a finite time horizon 
with a pay-off which depends upon the time the states and the control and then 
you get some pay-off from what happens at the end of your life so we have er a 
function phi here which depends upon where you end up so you can think of it as 
like being a prize or a reward or something so if you get big Xs here maybe 
this number's good if you get small Xs here maybe this number's bad but it's 
what the stock of assets of you own at the end of your life are actually worth 
to you in this problem now if we wanted to maximize this object we'd need to 
take account of the fact that 
where we end up has effects on what pay-off we're going to get from this 
problem so what is the relevant terminal condition in this problem well the 
relevant terminal condition is that lambda at time T is equal to the derivative 
of this function evaluated at X at time T so what does this say well suppose 
for some reason i was able to give you one more unit of final income one more 
unit of it how much more pay-off are you going to get well your pay-off is 
going to go up and find out it's close to this derivative so this says that 
lambda at time T has got to be equal to the marginal value to you of extra 
income at time T okay so this yeah that's what it says so you can think about 
this lambda at time T as being the value of extra X at time T 
nm0765: so we've been talking about these costate variables a lot we've talked 
about how we choose these costate variables and different constraints and how 
they affect the terminal value of these ker-, costate variables but i haven't 
really told you yet what these costate variables mean what they actually 
indicate so let's think about that lambda-T is the costate variable what is it 
well we know that it's a Lagrange multiplier and we know that Lagrange 
multipliers in some sense measure the cost of a constraint right when we look 
at straight consumer optimization problems the Lagrange multiplier there 
represents the marginal utility of mon-, money the Lagrange multiplier 
represents how much it costs the consumer to be constrained in its income if 
you could give the consumer more income the constraint would be weaker and the 
Lagrange multiplier measures that this 
is a Lagrange multiplier for a different sort of constraint it's essentially 
the Lagrange multiplier for the constraint that says X-prime of T is G-T- X-U 
right this is the constraint that relates X today to values of X tomorrow so 
what does this Lagrange multiplier measure well it measures the cost of this 
constraint how much having X linked in this way to today and tomorrow costs you 
as an optimizer now that doesn't seem particularly transparent until you write 
it in a different way lambda at time T measures the cost or benefit of a small 
increase in X at time T on on the solution to a an sorry optimal control 
problem so for some reason so what does lambda-T measure in an example think 
about er think about the problem of sending your rocket to the moon lambda at 
time T essentially measures maybe the costs associated with your fuel at time T 
so i s-, you're sending 
your your rocket to the moon and somehow someone comes along to you and says 
halfway to the moon i can give you another gallon of petrol how much is that 
worth to you well that's what lambda at time T measures the m-, the value to 
you somewhere along your problem of an extra unit of your state variable an 
extra unit of petrol an extra mile gone clear 
sf0768: what does lambda-T measure in your consumption problem 
nm0765: okay so it measures your money because what's the constraint in your 
consumption problem the constraint in your consumption problem was S-prime is 
equal to R-S-minus-C right so i can going to give you an extra unit's saving 
somewhere along the line so it measures the margin of utility of money to you 
sf0768: mm-hmm 
nm0765: yeah 
sf0768: mm-hmm 
nm0765: but it doesn't measure the marginal utility of money at any old point 
of time lambda at T repre-, represents the 
marginal utility of money to you at time T yeah fine er i can't think of a i 
can't think of a bath one but it would measure something like an extra egg cup 
full of water somewhere along the lay or er yeah er i think i've covered all 
the examples i understand if you can think of another one i'll have a go at it 
i've got another piece okay so we've done that let's say more about these 
costate variables 
nm0765: consider a problem with discounting now this is a some-, something that 
lots of mathematicians don't really understand but er let let's er we as 
economists do right this is a problem where we have a flow of utility from 
consumption at each point in time but this effect on our lifetime's income is 
discounted by the amount E-to-the-minus-R-T so one unit of consumption you to 
you at time zero is worth a lot le-, more to you than one unit of consumption 
at time 
capital-T because you're impatient and so this E-to-the-minus-R-T makes future 
consumption worth less to you than current consumption right fine so lambda at 
time T we've already talked about thanks to er namex measures the additional 
value to you of an extra unit of income at time T right so this is value to you 
of extra income at time little-T but that's not actually what you're interested 
in because you're now in period zero trying to decide what you're going to do 
for the whole of your life and lambda at time T measures how you feel about an 
extra unit of income at time little-T given you're currently at time zero so 
you might ask me the question you might if i gave how about my feeling about 
this extra unit of income at time T rather than at time little-zero well we 
define this thing called M at T to be E-to-the-power-R-T times lambda-T and 
what does that do well it scales up the the value of 
to you of income at time T by the discount factor E-to-the-R-T so it undoes the 
effect of the discounting and says to you oh let's let's let's step back what 
it does is says to you okay an extra unit of income to you at time T is worth 
this much to you at time zero to work out how much it's worth to you at time T 
we have to scale it up again 'cause when you get to time T the time T now is 
the present and so you have to undo the effect of the discounting so this thing 
here is called the the current marginal valuation now this thing here is going 
to be jolly useful for us in solving economic optimal control problems because 
in economic optimal control problems we often find that we have discounting so 
it's often okay it's often useful to eliminate the costate variable lambda at 
time T and replace it with M at time T the current marginal valuation so it's 
time for an example of that and we'll do that next time