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?


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

  • 1
    $\begingroup$ 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. $\endgroup$
    – Maxim
    Oct 4 '19 at 12:16
  • $\begingroup$ 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. $\endgroup$
    – Tom Mckay
    Oct 7 '19 at 13:32
  • 1
    $\begingroup$ 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). $\endgroup$
    – Maxim
    Oct 7 '19 at 14:34
  • $\begingroup$ Ok this was quite helpful. Does this mean that the integral then becomes $$\int_0^{\min(c, 1)}\int_0^{\min(c, a)} nz^{n - 1} dydz$$ $\endgroup$
    – Tom Mckay
    Oct 8 '19 at 11:57

$$\iint_{y<z<c}f_Y(y)f_Z(z)dydz=\iint_Dnz^{n-1}dydz$$ where $D=\{(z,y): 0<y<a \wedge 0<z<1 \wedge y<z<c\}$

It is best to do this with dz as the inner integral.

$$\int_0^{min(a,c)}\int_y^c nz^{n-1}dzdy=\int_0^{min(a,c)} c^n-y^n=min(a,c)c^n-\frac{min(a,c)^{n+1}}{n+1}$$

Therefore the expectation becomes

$$E(Z|Y_{n+1}<Z<c)=\frac{\iint_Dnz^{n}dydz}{\iint_Dnz^{n-1}dzdy}=\frac{\int_0^{min(a,c)} \frac{n}{n+1}(c^{n+1}-y^{n+1}dy)}{\int_0^{min(a,c)} c^n-y^ndy}$$ $$=\frac{min(a,c)\frac{n}{n+1}c^{n+1}-\frac{n}{(n+1)(n+2)}min(a,c)^{n+2}}{min(a,c)c^n-\frac{min(a,c)^{n+1}}{n+1}}$$ which is the same as the previous result whenever $min(a,c)=c$

Much thanks to Maxim for the hints.


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