Stochastic Calculus

[WMD]

Conditional Expectation:

Suppose we roll two dice, a red and a green one, and let X be the value on the red die and Y the value on the green die. Let Z = X/Y

Find E[ (X+2Y) | Z ]

My answer:

[math]E[ZY| ] =\displaystyle\sum_{Y=1}^6\displaystyle\sum_{X=1}^6 \frac{X}{Y}\cdot{Y}\cdot \frac{1}{36}= \frac72[/math]

[math]E[Z] =\displaystyle\sum_{Y=1}^6\displaystyle\sum_{X=1}^6 \frac{X}{Y}\cdot \frac{1}{36} = 1.4291667 \therefore E[Y|Z] =\frac{E[ZY]}{E[Z]} = \frac{\frac72}{1.4291667}=2.448979 \therefore E[(X+ 2Y)|Z] = (Z + 2) \cdot 2.448979[/math]
Is the above answer correct?

 

[MikeLawrence]

I don't think it is. Also, what part of this is stochastic calculus? If this is stochastic calculus, then I covered a whole lot of it in my mathematical statistics courses.

Isn't the answer of the first bit 0.5833333, and not 7/2? The Y's cancel, so you're left with X(1/36) summed from x = 1:6. So.. (1+2+3+4+5+6)/36 = 0.583333.

Anyone else have thoughts? Good to see a technical question on here. This should happen more often. If there is a thread for this I haven't seen, someone please link it. I saw one post about reducing a fraction or something that was interesting.

 

[WMD]

I don't think it is. Also, what part of this is stochastic calculus? If this is stochastic calculus, then I covered a whole lot of it in my mathematical statistics courses.

Isn't the answer of the first bit 0.5833333, and not 7/2? The Y's cancel, so you're left with X(1/36) summed from x = 1:6. So.. (1+2+3+4+5+6)/36 = 0.583333.

Anyone else have thoughts? Good to see a technical question on here. This should happen more often. If there is a thread for this I haven't seen, someone please link it. I saw one post about reducing a fraction or something that was interesting.

If we change the probability, We will get your answer. [math]E[ZY]=\displaystyle\sum_{Y=1}^6\displaystyle\sum_{X=1}^6 \frac{X}{Y}\cdot Y\cdot \frac{1}{216}= 0.58333333 \therefore E[Y|Z] =\frac{0.583333}{1.4291667}= 0.4081632324[/math]
This answer may be correct.

 

[MikeLawrence]

You misunderstood. I didn't sum over any of the Y's in the first equation. They divide into each other and leave 1, so you only sum the X's. The Y portion disappears. So it is just the sum of X from X = 1:6 of (X/36).

 

[WMD]

So, as probability of X is [math]\frac16[/math], we have [math]E[Z \cdot Y]=\displaystyle\sum_{X=1}^6 X\cdot\frac16 =3.5[/math] So, my answer given in #1 is correct.

How would you answer the following question?

Let W = E[Z | X]. What are the possible values for W? Give the distribution of W.

My answer: Possible values of W= E[Z| X] are

W = E[Z | X = 1] = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6)/6 = 0.4083...
W = E[Z | X = 2] = (2 + 1 + 2/3 + 1/2 + 2/5 + 1/3)/6 = 0.8166...
W = E[Z | X = 3] = (3 + 3/2 + 1 + 3/4 + 3/5 + 1/2)/6 = 1.225
W = E[Z | X = 4] = (4 + 2 + 4/3 + 1 + 4/5 + 2/3)/6 = 1.6333...
W = E[Z | X = 5] = (5 + 5/2 + 5/3 + 5/4 + 1 + 5/6)/6 = 2.0416...
W = E[Z | X = 6] = (6 + 3 + 2 + 3/2 + 6/5 + 1)6 = 2.45.

How would you compute the distribution of W?

 

[GLawler]

This problem comes from the exercises in my book on Stochastic Calculus. It is in the first section where the definition of a conditional expectation is given. The intention was just to work it out (a tedious exercise) over all the values that Z can take. The purpose of the exercise is to reinforce the definition of a conditional expectation.

There is no "formula for conditional expectation" of the type you gave --- you are giving a number and the conditional expectation should be a random variable measurable with respect to Z. For a discrete random variable, it is the same as giving for each value Z = z , the conditional expectation given this value.

 

[GLawler]

How would you answer the following question?

Let W = E[Z | X]. What are the possible values for W? Give the distribution of W.

My answer: Possible values of W= E[Z| X] are

W = E[Z | X = 1] = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6)/6 = 0.4083...
W = E[Z | X = 2] = (2 + 1 + 2/3 + 1/2 + 2/5 + 1/3)/6 = 0.8166...
W = E[Z | X = 3] = (3 + 3/2 + 1 + 3/4 + 3/5 + 1/2)/6 = 1.225
W = E[Z | X = 4] = (4 + 2 + 4/3 + 1 + 4/5 + 2/3)/6 = 1.6333...
W = E[Z | X = 5] = (5 + 5/2 + 5/3 + 5/4 + 1 + 5/6)/6 = 2.0416...
W = E[Z | X = 6] = (6 + 3 + 2 + 3/2 + 6/5 + 1)6 = 2.45.

How would you compute the distribution of W?

Each of the six values you gave happens with probability 1/6. That gives the distribution of W. For example, P(W = 0.4083...) = 1/6.

Again, this is an exercise to learn the meaning of the terms.

 

[MikeLawrence]

This problem comes from the exercises in my book on Stochastic Calculus. It is in the first section where the definition of a conditional expectation is given. The intention was just to work it out (a tedious exercise) over all the values that Z can take. The purpose of the exercise is to reinforce the definition of a conditional expectation.

There is no "formula for conditional expectation" of the type you gave --- you are giving a number and the conditional expectation should be a random variable measurable with respect to Z. For a discrete random variable, it is the same as giving for each value Z = z , the conditional expectation given this value.

Thanks for the answer. I think your response only pertains to the post directly above yours, do you have anything to say about the original post?

 

[GLawler]

My first response was to the original post

 

[MikeLawrence]

My first response was to the original post

Oh, I think I get it now. I forgot OP had computed a conditional expectation in the original post.

 

[Andy Nguyen]

This problem comes from the exercises in my book on Stochastic Calculus.

Prof. Lawler, very nice to see you here and thank you for post.
If you don't mind, I'll post the link to your page on University of Chicago's Financial Mathematics program which contains link to your Stochastic Calculus ebook.

Gregory Lawler | Financial Mathematics | The University of Chicago

The University of Chicago’s Financial Mathematics Program offers courses in option pricing, portfolio management, machine learning, and python to prepare students for careers in quantitative finance.

finmath.uchicago.edu

 

[WMD]

This problem comes from the exercises in my book on Stochastic Calculus. It is in the first section where the definition of a conditional expectation is given. The intention was just to work it out (a tedious exercise) over all the values that Z can take. The purpose of the exercise is to reinforce the definition of a conditional expectation.

There is no "formula for conditional expectation" of the type you gave --- you are giving a number and the conditional expectation should be a random variable measurable with respect to Z. For a discrete random variable, it is the same as giving for each value Z = z , the conditional expectation given this value.

Hello,
Professor, I am happy to talk with you here.

Professor G. Lawler, I haven't read your book yet completely. I am trying to solve the exercises given by you in your book. But the drawback of your book is that there is no 'answers to exercises' section.

I recommend/request you to provide ' answers to exercises' section in your book so that students would do their homework easily and simultaneously verify their work with 'answers to exercises' section. You commented on my answer but I understood it vaguely.

So, I repeat my question. Is my answer to E[ (X+2y) |Z ] correct?

 

[GLawler]

I think there are others on this forum who should now be able to help you understand the problems with your answer given what I already posted. I mainly posted because of the comment as to why it was from a stochastic calculus book. I am happy to have a link to my notes posted, but I cannot serve as an individual tutor for people on the web. My priority for time to explain mathematics must be for students enrolled at University of Chicago.

For your information, students at University of Chicago are provided solutions to homework exercises after they have been turned in.

 

[WMD]

You misunderstood. I didn't sum over any of the Y's in the first equation. They divide into each other and leave 1, so you only sum the X's. The Y portion disappears. So it is just the sum of X from X = 1:6 of (X/36).

Following are the two tables, I prepared for answering this question.

Considering these above tables, we can ascertain the expected value of (X +2Y) given Z.
E[ X + 2Y | Z] = [3 + 6 + 9 + 12 + 15+ 18 =63]* 1/6 = 10.5 + [5 +10+15]*1/12= 2.5 + [7+ 14]*1/18 =1.1667 + [9 + 11 + 13 +12 + 11+ 13+ 6 + 10 +14 + 7 +9 + 11 +13 +17 + 8 +16 =179]*1/36 = 4.9722222 +[4 +8 +12=24]*1/12=2 + [ 8 +16=24]*1/18 = 1.33333 +[5 + 10 = 15]*1/18 =0.8333333 + [7 +14=21]*1/18 = 1.1667 = 24.47223