Suppose that $Y_1, Y_2$ are two independent observations from the following distribution: $$ f_\theta(y) = \begin{cases} \dfrac{3y^2}{\theta^3} &\text{for } 0 <y \leq \theta, \ \theta>0 \\[8pt] 0 & \text{otherwise}\end{cases} $$

I am trying to find $E(Y_1 \mid \max(Y_1, Y_2))$, and am not sure how to do this. I can directly compute the integral but I still need to find the joint distribution of $Y_1$ and $\max(Y_1,Y_2)$, which I am not sure how to derive.

  • $\begingroup$ your pdf does not integrate to 1 isn't it $\frac{3y^2}{\theta^3}$? $\endgroup$ – Momo Dec 5 '16 at 3:00
  • $\begingroup$ Yes it is, sorry for the oversight! $\endgroup$ – user321627 Dec 5 '16 at 3:09

Without computing the joint density:

Let's define an indicator random variable $K$ taking value $1$ if $Y_1 > Y_2$, $2$ otherwise. Clearly, from symmetry, $P(K=1)=\frac12$. Let $Z=\max(Y_1,Y_2)$.

Then, using the law of total expectation with respect to variable $K$, $$E(Y_1 \mid Z) =E( E(Y_1 \mid Z , K))=\frac12 E(Y_1 \mid Z, K=1)+\frac12 E(Y_1 \mid Z ,K=2)$$

But $E(Y_1 \mid Z, K=1)=Z$ and $$E(Y_1 \mid Z, K=2)=E(Y_1 \mid Y_1 < Z)$$

This expectation corresponds to that of a truncated $Y_1$:

$$E(Y_1 \mid Y_1 < Z)=\frac{\int_0^Z y\, 3 y^2\, dy }{\int_0^Z 3 y^2 \,dy}=\frac{3}{4}Z$$

Putting all together:

$$E(Y_1 \mid Z)=\frac{1}{2}Z+\frac{3}{8}Z=\frac{7}{8}Z$$

  • $\begingroup$ Thanks! If I could ask, could you tell me how you got $E( E(Y_1 \mid Z , K))=\frac12 E(Y_1 \mid Z, K=1)+\frac12 E(Y_1 \mid Z ,K=2)$? It seems like the expectation is taken with respect to $Z$ and $K$, but I am not sure how the right hand side is derived. $\endgroup$ – user321627 Dec 5 '16 at 3:13
  • $\begingroup$ @user321627 Instead of $K=1$ read $Y_1>Y_2$... $\endgroup$ – Momo Dec 5 '16 at 3:27
  • $\begingroup$ Can I also ask how you got $E(Y_1 \mid Z, K=1)=Z$ and $E(Y_1 \mid Z K=2)=E(Y_1 \mid Y_1 < Z)$? $\endgroup$ – user321627 Dec 5 '16 at 4:11
  • $\begingroup$ @user321627 In $E(E(Y1∣Z,K))$ the outer expectation is with respect of $K$ (total expectation), so I'm computing that expectation using the usual $E(g(X))=\sum p(x)g(x)$ $\endgroup$ – leonbloy Dec 5 '16 at 4:26
  • $\begingroup$ $E(Y_1 \mid Z, K=1)=Z$ because knowing that $K=1$ amount to knowing that $Z=Y_1$ $\endgroup$ – leonbloy Dec 5 '16 at 4:27

Hint: to find the joint pdf of $(Y_1,\max(Y_1,Y_2))$:

The support of $(Y_1,\max(Y_1,Y_2))$ is

Observe first that the CDF of $Y_i$ is $F_\theta(y)=\frac{y^3}{\theta^3}$ for $ 0<y<\theta$

Now if $F$ and $f$ are the joint CDF and pdf of $Y_1$ and $Y_2$, then for $0<y_1, y_2\le\theta$:

$$F(y_1,y_2)=P(Y_1\le y_1,\max(Y_1,Y_2)\le y_2)=P(Y_1\le y_1,Y_1\le y_2,Y_2\le y_2)\\=P(Y_1\le\min(y_1,y_2),Y_2\le y_2)=P(Y_1\le \min(y_1,y_2))P(Y_2\le y_2)=F_\theta(\min(y_1,y_2))F_\theta(y_2)$$


$$F(y_1,y_2)= \begin{cases} F_\theta(y_1)F_\theta(y_2)=\frac{9 y_1^2 y_2^2}{\theta^6}& \text{ if } y_1\le y_2\\ F_\theta(y_2)F_\theta(y_2)=\frac{9 y_2^4}{\theta^6}& \text{ if } y_1> y_2 \end{cases} $$

The problem with this approach is that when you differentiate, there is a Dirac component coming from a discontinuity of $F$, which is messy to work with.

So for the simplest path to final goal, you should follow leonbloy advice in his answer and condition on whether $Y_1\le Y_2$ or not:

$$E[Y_1|\max(Y_1,Y_2)]=E[Y_1|\max(Y_1,Y_2),Y_1\le Y_2]P(Y_1\le Y_2)+E[Y_1|\max(Y_1,Y_2),Y_1>Y_2]P(Y_1>Y_2)$$

  • $\begingroup$ Thanks, can you tell me why you chose to have $y_1 \leq y_2$? $\endgroup$ – user321627 Dec 5 '16 at 2:34
  • $\begingroup$ because $Y_1\le\max(Y_1,Y_2)$ so the joint pdf is zero if $y_1>y_2$ $\endgroup$ – Momo Dec 5 '16 at 2:35
  • $\begingroup$ could you elaborate on why the joint pdf would be zero? $\endgroup$ – user321627 Dec 5 '16 at 2:37
  • $\begingroup$ yeah, I think you are right. let me edit. $\endgroup$ – Momo Dec 5 '16 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.