Conditional expectation $E[\min{\{X,Y\}}|Y^{2}]$

Let $X$ and $Y$ be an independent random variables with exponential distribution $Exp(1)$.

Compute $E[\min{\{X,Y\}}|Y^{2}]$.

My attempt:

I. If we want to compute $E[\min{\{X,Y\}}|Y]$ may be sufficient to use the following formula:

$$E[f(X,Y)|Y]=\int\limits_{-\infty}^{+\infty}f(x,Y)f_{X}(x)dx$$

Is this correct? Is there any other simpler method to compute this conditional expectation?

II. What about $E[\min{\{X,Y\}}|Y^{2}]$? First thing which comes to my mind is that we should find distribution of vector $(\min{\{X,Y\}},Y^{2})$ and $Y^{2}$ and then use appropriate formula. Am I right? How to find distribution of vector $(\min{\{X,Y\}},Y^{2})$? Is there any simpler method to compute this conditional expecation?

• HINT: if $X \sim Exp(\lambda_x)$ and $Y \sim Exp(\lambda_y)$, then $Z = \min\{X,Y\} \sim Exp(\lambda_x + \lambda_y)$ – Guilherme Thompson Dec 14 '15 at 9:11
• @GuilhermeThompson ?? "HINT" is not useful and "HINT2" is wrong. – Did Mar 19 '16 at 10:43

1. Yes, you are correct. So we have

$$\Bbb{E}[\min\{X, Y\} | Y] = \int_{0}^{\infty} \min\{x, Y\} \, e^{-x} \, dx = 1 - e^{-Y}.$$

2. Notice that both $Y$ and $Y^2$ generate the same $\sigma$-algebra: $\sigma(Y) = \sigma(Y^2)$. So we have

$$\Bbb{E}[f(X, Y)|Y^2] = \Bbb{E}[f(X, Y)|\sigma(Y^2)] = \Bbb{E}[f(X, Y)|\sigma(Y)] = \Bbb{E}[f(X, Y)|Y].$$

If you have not learned the axiomatic approach to the probability theory, ignore this explanation. We have an alternative, more computational explanation as follows: Let $Z = Y^2$. Then

\begin{align*} \Bbb{E}[f(X, Y) | Y^2] &= \Bbb{E}[f(X, \sqrt{Z}) | Z] \\ &= \int_{0}^{\infty} f(x, \sqrt{Z}) \, e^{-x} \, dx \\ &= \int_{0}^{\infty} f(x, Y) \, e^{-x} \, dx \\ &= \Bbb{E}[f(X, Y) | Y]. \end{align*}

• @Leon, Yes, any exponential r.v. is non-negative with probability 1. – Sangchul Lee Dec 14 '15 at 9:50
• I am stuck at a similar problem, would you mind adding a comment on what you did to compute the integral? i.e. $$\int_0^\infty \min \lbrace x,Y \rbrace e^{-x} dx = ?$$ – Spaced Oct 1 '16 at 18:34
• @Spaced, Have you tried splitting the integral into two parts according to whether $x < Y$ or $x > Y$? – Sangchul Lee Oct 1 '16 at 19:58
• Thanks a lot that did the trick! (the best thing being that it's not even a trick, more like an application of the indicator function) – Spaced Oct 1 '16 at 20:49