Calculate conditional expectation of exponential variables

In summary, the joint density for X and Y is abe^(-ax)e^(-by) for x,y>0, and the marginal density for w is abe^(-bw) * (e^(bw-aw) - 1) / (b-a). The conditional density for (Z|W=w) is e^(bz-az) * (b-a) / (e^(bw-aw) - 1). After integrating from 0 to w, the simplified expression for E(X|X+Y) is (e^(cw)(cw-1)+1) / (c(e^(cw)-1)), where c = b-a.
  • #1
obst12
3
0

Homework Statement


Let X and Y be independent exponential random variables with parameters a and b. Calculate E(X|X+Y).

Homework Equations

The Attempt at a Solution


I'm pretty sure I have it, just want to make sure.

Joint density for X and Y is abe^(-ax)e^(-by) for x,y>0. Let Z=X and W=X+Y so X=Z and Y=W-Z. The Jacobian is 1, so the joint density is

abe^(-az)e^(-bw+bz)=abe^(+bz-az) e^(-bw)

Marginal density for w is

\int_0^ w abe^(bz-az) e^(-bw) dz=ab(e^(bw-aw)-1) /(b-a)

So then conditional density is

(b-a)e^(bz-az) e^(-bw)/(e^(bw-aw)-1)

Then just integrate z from 0 to infinity to get
e^(-bw)/(1-e^(bw-aw))

Is this okay?
 
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  • #2
obst12 said:

Homework Statement


Let X and Y be independent exponential random variables with parameters a and b. Calculate E(X|X+Y).

Homework Equations

The Attempt at a Solution


I'm pretty sure I have it, just want to make sure.

Joint density for X and Y is abe^(-ax)e^(-by) for x,y>0. Let Z=X and W=X+Y so X=Z and Y=W-Z. The Jacobian is 1, so the joint density is

abe^(-az)e^(-bw+bz)=abe^(+bz-az) e^(-bw)

Marginal density for w is

\int_0^ w abe^(bz-az) e^(-bw) dz=ab(e^(bw-aw)-1) /(b-a)

So then conditional density is

(b-a)e^(bz-az) e^(-bw)/(e^(bw-aw)-1)

Then just integrate z from 0 to infinity to get
e^(-bw)/(1-e^(bw-aw))

Is this okay?

No, it is not OK. Given ##X+Y = w##, ##X## has range ##[0,w]##; in your notation, ##0 \leq (Z|W=w) \leq w##. Therefore, integrating out to ##\infty## is wrong. Also, your "conditional density" ##f(z|w)## for ##(Z|W=w)## looks incorrect a bit. Go back to square one and proceed very carefully. The correct final answer is more complicated than yours.
 
  • #3
Thanks. Let me try once more.

The joint density is
##abe^{-ax}e^{-by}##
Then making the substitution ##X=Z## and ##Y=W-Z##.
##\rho_{Z.W}=abe^{-az}e^{-bw+bz}##
##\rho_{Z,W}=abe^{(b-a)z}e^{-bw}##

Thus the marginal density is
##\rho_W=\int_0^w abe^{(b-a)z}e^{-bw} dz##
##\rho_W=abe^{-bw} \int_0^w e^{(b-a)z} dz##
##rho_W=abe^{-bw} \frac{1}{b-a} (e^{(b-a)w}-1)##

And the conditional density is
##\rho_{Z|W}=abe^{(b-a)z}e^{-bw}\frac{1}{abe^{-bw} \frac{1}{b-a} (e^{(b-a)w}-1)}##
##\rho_{Z|W}=e^{(b-a)z}\frac{1}{\frac{1}{b-a} (e^{(b-a)w}-1)}##
##\rho_{Z|W}=e^{(b-a)z}\frac{b-a}{(e^{(b-a)w}-1)}##

Then integrate from ##0## to ##w##

##\mathbb{E}(Z|W)=\int_0^w z e^{(b-a)z}\frac{b-a}{(e^{(b-a)w}-1)} dz##
##\mathbb{E}(Z|W)=\frac{b-a}{(e^{(b-a)w}-1)} \int_0^w z e^{(b-a)z} dz##
##\mathbb{E}(Z|W)=\frac{b-a}{(e^{(b-a)w}-1)} \frac{e^{(b-a)w}((b-a)w-1)+1}{(b-a)^2}##

Is this okay? Thanks for your help.
 
  • #4
obst12 said:
Thanks. Let me try once more.

The joint density is
##abe^{-ax}e^{-by}##
Then making the substitution ##X=Z## and ##Y=W-Z##.
##\rho_{Z.W}=abe^{-az}e^{-bw+bz}##
##\rho_{Z,W}=abe^{(b-a)z}e^{-bw}##

Thus the marginal density is
##\rho_W=\int_0^w abe^{(b-a)z}e^{-bw} dz##
##\rho_W=abe^{-bw} \int_0^w e^{(b-a)z} dz##
##rho_W=abe^{-bw} \frac{1}{b-a} (e^{(b-a)w}-1)##

And the conditional density is
##\rho_{Z|W}=abe^{(b-a)z}e^{-bw}\frac{1}{abe^{-bw} \frac{1}{b-a} (e^{(b-a)w}-1)}##
##\rho_{Z|W}=e^{(b-a)z}\frac{1}{\frac{1}{b-a} (e^{(b-a)w}-1)}##
##\rho_{Z|W}=e^{(b-a)z}\frac{b-a}{(e^{(b-a)w}-1)}##

Then integrate from ##0## to ##w##

##\mathbb{E}(Z|W)=\int_0^w z e^{(b-a)z}\frac{b-a}{(e^{(b-a)w}-1)} dz##
##\mathbb{E}(Z|W)=\frac{b-a}{(e^{(b-a)w}-1)} \int_0^w z e^{(b-a)z} dz##
##\mathbb{E}(Z|W)=\frac{b-a}{(e^{(b-a)w}-1)} \frac{e^{(b-a)w}((b-a)w-1)+1}{(b-a)^2}##

Is this okay? Thanks for your help.

Yes, it's OK, but it should be simplified a bit. You can make it look nicer by writing
[tex] E(X|X+Y=w) = \frac{e^{cw}(cw-1)+1}{c(e^{cw}-1)}, \: c = b-a[/tex]
 
  • #5
Awesome, thanks for all your help!
 

1. What is conditional expectation of exponential variables?

Conditional expectation of exponential variables is a statistical concept that calculates the expected value of an exponential variable given a specific event or condition. It is a measure of the average outcome of an exponential variable when that variable is restricted to a certain subset of all possible outcomes.

2. How is conditional expectation of exponential variables calculated?

The formula for calculating conditional expectation of exponential variables is E[X|A] = ∫A xf(x) / ∫A f(x), where X is the exponential variable, A is the event or condition, and f(x) is the probability density function of X. This formula follows the general definition of conditional expectation, which is the expected value of X given A is true.

3. What is the significance of calculating conditional expectation of exponential variables?

Conditional expectation of exponential variables has many practical applications in statistics and decision-making. It can help in predicting future outcomes based on past data, evaluating the effectiveness of a certain strategy or treatment, and determining the best course of action in uncertain situations.

4. Can conditional expectation of exponential variables be negative?

Yes, it is possible for the conditional expectation of exponential variables to be negative. This occurs when the event A has a high probability and the exponential variable X has a low expected value. In this case, the expected value of X given A is true will be negative.

5. How does conditional expectation of exponential variables differ from unconditional expectation?

Unconditional expectation is the expected value of an exponential variable without any restrictions or conditions. It takes into account all possible outcomes of the variable. On the other hand, conditional expectation only considers a subset of outcomes, based on a given event or condition. This makes the value of conditional expectation different from unconditional expectation in most cases.

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