Conditional expected value of a maximum of uniform random variables with differing supports.

This is an extension of the previous question here.

Conditional expected value of a maximum of uniform random variables

I have $$X_{1}$$,$$X_{2}$$... $$X_{n}$$ independent uniform random variables on $$[0,1]$$.

I also have $$Y_{n+1}$$ which is a uniform distribution on $$[0,a]$$ where $$a\in [0,1]$$

let $$Z=max(X_{1},X_{2}... X_{n})$$

let $$c$$ be a constant s.t $$c \in [0,1]$$

What is the following conditional expectation?

$$E(Z|Y_{n+1}

The previous post contains a solution for when $$a=1$$ but I am unsure how to proceed when the supports are different.

• Which step are you having a difficulty with? To convert the double integrals to iterated integrals, draw the region $0 < y < a \land 0 < z < 1 \land y < z < c$ in the $(z, y)$ plane. Oct 4 '19 at 12:16
• Im sorry im not exactly sure where to start. Using your method from before I get to the double integral and get stuck. Let $c > 0$. We have $f_Y(x) = [0 < x < a], \, f_Z(x) = n x^{n - 1} [0 < x < 1]$, $$\operatorname{E}(Z \mid Y < Z < c) = \frac {\operatorname{E}(Z \, [Y < Z < c])} {\operatorname{P}(Y < Z < c)} = \\ \frac {\iint_{y < z < c} z f_Y(x) f_Z(z) \, dx dz} {\iint_{y < z < c} f_Y(x) f_Z(z) \, dx dz}$$ Where does a enter the following integral, my intuition is that I would have to use Identity functions but there must be a better solution. Oct 7 '19 at 13:32
• We have $$\iint_{y < z < c} f_Y(y) f_Z(z) \, dy dz = \iint_D n z^{n - 1} dy dz, \\ D = \{(z, y): 0 < y < a \land 0 < z < 1 \land y < z < c\}.$$ Can you visualize the set of points $(z, y)$ comprising $D$? Once you do that, it should be clear why $D$ is a normal domain and how to find the functions $\alpha$ and $\beta$ (and which axis it's more convenient to take the projection on). Oct 7 '19 at 14:34
• Ok this was quite helpful. Does this mean that the integral then becomes $$\int_0^{\min(c, 1)}\int_0^{\min(c, a)} nz^{n - 1} dydz$$ Oct 8 '19 at 11:57

$$\iint_{y where $$D=\{(z,y): 0
$$\int_0^{min(a,c)}\int_y^c nz^{n-1}dzdy=\int_0^{min(a,c)} c^n-y^n=min(a,c)c^n-\frac{min(a,c)^{n+1}}{n+1}$$
$$E(Z|Y_{n+1} $$=\frac{min(a,c)\frac{n}{n+1}c^{n+1}-\frac{n}{(n+1)(n+2)}min(a,c)^{n+2}}{min(a,c)c^n-\frac{min(a,c)^{n+1}}{n+1}}$$ which is the same as the previous result whenever $$min(a,c)=c$$