Question about book solution to estimate $E[e^{XY}]$ when $X$ and $Y$ are independent exponential RVs with...
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Let $X$ and $Y$ be independent exponential random variables with mean 1. (a) Explain how we could use simulation to estimate $E[e^{XY}]$. (b) Show how to improve the estimation approach in part (a) by using a control variate.
The red box (see screenshot) in the book solution makes sense to me.
I don't understand the blue box. It seems to me the blue box estimator will not have the same expected value as $E[e^{XY}]$ and is missing a "$-c$" in the numerator because $E[X_i Y_i] = E[X_i] E[Y_i] = 1$
So I think the blue box should actually be
$$sum_{i=1}^n frac{e^{X_i Y_i} + c X_i Y_i - c}{n}$$
Will the blue box estimator have the same expected value as $E[e^{XY}]$? Thanks for your help I'm learning about control variates for the first time.
Book Solution
probability law-of-large-numbers expected-value
asked Nov 17 at 18:41
HJ_beginner
855315
add a comment |
up vote
1
down vote
favorite
Let $X$ and $Y$ be independent exponential random variables with mean 1. (a) Explain how we could use simulation to estimate $E[e^{XY}]$. (b) Show how to improve the estimation approach in part (a) by using a control variate.
The red box (see screenshot) in the book solution makes sense to me.
I don't understand the blue box. It seems to me the blue box estimator will not have the same expected value as $E[e^{XY}]$ and is missing a "$-c$" in the numerator because $E[X_i Y_i] = E[X_i] E[Y_i] = 1$
So I think the blue box should actually be
$$sum_{i=1}^n frac{e^{X_i Y_i} + c X_i Y_i - c}{n}$$
Will the blue box estimator have the same expected value as $E[e^{XY}]$? Thanks for your help I'm learning about control variates for the first time.
Book Solution
probability law-of-large-numbers expected-value
asked Nov 17 at 18:41
HJ_beginner
855315
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ and $Y$ be independent exponential random variables with mean 1. (a) Explain how we could use simulation to estimate $E[e^{XY}]$. (b) Show how to improve the estimation approach in part (a) by using a control variate.
The red box (see screenshot) in the book solution makes sense to me.
I don't understand the blue box. It seems to me the blue box estimator will not have the same expected value as $E[e^{XY}]$ and is missing a "$-c$" in the numerator because $E[X_i Y_i] = E[X_i] E[Y_i] = 1$
So I think the blue box should actually be
$$sum_{i=1}^n frac{e^{X_i Y_i} + c X_i Y_i - c}{n}$$
Will the blue box estimator have the same expected value as $E[e^{XY}]$? Thanks for your help I'm learning about control variates for the first time.
Book Solution
probability law-of-large-numbers expected-value
asked Nov 17 at 18:41
HJ_beginner
855315
Let $X$ and $Y$ be independent exponential random variables with mean 1. (a) Explain how we could use simulation to estimate $E[e^{XY}]$. (b) Show how to improve the estimation approach in part (a) by using a control variate.
The red box (see screenshot) in the book solution makes sense to me.
I don't understand the blue box. It seems to me the blue box estimator will not have the same expected value as $E[e^{XY}]$ and is missing a "$-c$" in the numerator because $E[X_i Y_i] = E[X_i] E[Y_i] = 1$
So I think the blue box should actually be
$$sum_{i=1}^n frac{e^{X_i Y_i} + c X_i Y_i - c}{n}$$
Will the blue box estimator have the same expected value as $E[e^{XY}]$? Thanks for your help I'm learning about control variates for the first time.
Book Solution
probability law-of-large-numbers expected-value
probability law-of-large-numbers expected-value
asked Nov 17 at 18:41
HJ_beginner
855315
asked Nov 17 at 18:41
HJ_beginner
855315
asked Nov 17 at 18:41
HJ_beginner
855315
asked Nov 17 at 18:41
HJ_beginner
855315
asked Nov 17 at 18:41
HJ_beginner
855315
855315
add a comment |
add a comment |
1 Answer
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Yes, for $c in mathbb{R}$ the first should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i-1)}_{mbox{mean 0}}]$$ and also the second should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i + X_i^2Y_i^2/2 - 3)}_{mbox{mean 0}}]$$
Using $c=-1$ seems to be a good choice for reducing variance.
answered Nov 17 at 19:57
Michael
13.1k11325
Thanks for your help Michael! This was really bothering me, especially as it is the final question of the book that has taken me 2 years of self study to get through lol fml (fantastic book by Ross, I'm just slow)
– HJ_beginner
Nov 17 at 20:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, for $c in mathbb{R}$ the first should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i-1)}_{mbox{mean 0}}]$$ and also the second should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i + X_i^2Y_i^2/2 - 3)}_{mbox{mean 0}}]$$
Using $c=-1$ seems to be a good choice for reducing variance.
answered Nov 17 at 19:57
Michael
13.1k11325
Thanks for your help Michael! This was really bothering me, especially as it is the final question of the book that has taken me 2 years of self study to get through lol fml (fantastic book by Ross, I'm just slow)
– HJ_beginner
Nov 17 at 20:32
add a comment |
up vote
1
down vote
accepted
Yes, for $c in mathbb{R}$ the first should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i-1)}_{mbox{mean 0}}]$$ and also the second should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i + X_i^2Y_i^2/2 - 3)}_{mbox{mean 0}}]$$
Using $c=-1$ seems to be a good choice for reducing variance.
answered Nov 17 at 19:57
Michael
13.1k11325
Thanks for your help Michael! This was really bothering me, especially as it is the final question of the book that has taken me 2 years of self study to get through lol fml (fantastic book by Ross, I'm just slow)
– HJ_beginner
Nov 17 at 20:32
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, for $c in mathbb{R}$ the first should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i-1)}_{mbox{mean 0}}]$$ and also the second should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i + X_i^2Y_i^2/2 - 3)}_{mbox{mean 0}}]$$
Using $c=-1$ seems to be a good choice for reducing variance.
answered Nov 17 at 19:57
Michael
13.1k11325
Yes, for $c in mathbb{R}$ the first should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i-1)}_{mbox{mean 0}}]$$ and also the second should be $$frac{1}{n}sum_{i=1}^n[e^{X_iY_i} + cunderbrace{(X_iY_i + X_i^2Y_i^2/2 - 3)}_{mbox{mean 0}}]$$
Using $c=-1$ seems to be a good choice for reducing variance.
answered Nov 17 at 19:57
Michael
13.1k11325
answered Nov 17 at 19:57
Michael
13.1k11325
answered Nov 17 at 19:57
Michael
13.1k11325
answered Nov 17 at 19:57
Michael
13.1k11325
13.1k11325
Thanks for your help Michael! This was really bothering me, especially as it is the final question of the book that has taken me 2 years of self study to get through lol fml (fantastic book by Ross, I'm just slow)
– HJ_beginner
Nov 17 at 20:32
add a comment |
Thanks for your help Michael! This was really bothering me, especially as it is the final question of the book that has taken me 2 years of self study to get through lol fml (fantastic book by Ross, I'm just slow)
– HJ_beginner
Nov 17 at 20:32
Thanks for your help Michael! This was really bothering me, especially as it is the final question of the book that has taken me 2 years of self study to get through lol fml (fantastic book by Ross, I'm just slow)
– HJ_beginner
Nov 17 at 20:32
Thanks for your help Michael! This was really bothering me, especially as it is the final question of the book that has taken me 2 years of self study to get through lol fml (fantastic book by Ross, I'm just slow)
– HJ_beginner
Nov 17 at 20:32
add a comment |
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