Calculate $E(max(A, B))$

Question

If I got two independent random variables $A \sim R(0,1)$ and $B \sim R(0,1)$ and I have to calculate $E(max(A, B))$ wouldn't that then just be to pick one of them since they are equal in size? So it would be the same $E(A)$?

What would E(max{A,B}) be?

No. Realize that $A\leq\max(A,B)$ and secondly that $\Pr(A<\max(A,B))>0$. This is enough to conclude that $\mathbb EA<\mathbb E\max(A,B)$. — drhab, Mar 27 '17 at 08:52
@AndrewC That's another question. If you want an anwer to that add it to your original question with also your own efforts to solve it. — drhab, Mar 27 '17 at 09:25

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

To compute $E(max(A, B))$, I'd start by computing $P(max(A, B) \le x)$. As Alain Chau has observed, $P(max(A, B) \le x) = P(A \le x) P(B \le x)$ since $A$ and $B$ are independent.

Since $A$ and $B$ are uniformly distributed on $[0, 1]$, you have $P(max(A, B) \le x) = x^2$ for $0 \le x \le 1$.

Finally, use the well-known formula $\int_0^\infty P(X > x) \: dx = E(X)$, which applies for non-negative random variables $X$ (this question may be relevant).

score 1 · Answer 2 · answered Mar 27 '17 at 09:33

To aid your intuition, consider the expected value of a fair die versus the expected value of rolling two dice and taking their maximum; it should be clear that their expected values are not the same. Mathematically, you can get around the max function by observing that $$P(\max(A,B)\leq x) = P(A \leq x)\cdot P(B\leq x) $$ since $$\max(A,B) \leq x \iff A \leq x \text{ and } B\leq x$$ and A and B are independent.

drhab · Accepted Answer · 2017-03-27T09:51:17.100

0

Let $\epsilon>0$ and let $E$ denote the event that $A+\epsilon\leq B$.

Then: $$(A+\epsilon)1_E\leq\max(A,B)1_E$$ and consequently: $$\mathbb EA1_{E}+\epsilon\Pr(E)\leq\mathbb E\max(A,B)1_E$$

Taking $\epsilon>0$ small enough we get $\Pr(E)>0$ so that:

$$\mathbb EA1_E<\mathbb EA1_{E}+\epsilon\Pr(E)$$

That leads to:

$$\mathbb EA=\mathbb EA1_E+\mathbb EA1_{E^c}<E\max(A,B)1_E+E\max(A,B)1_{E^c}=\mathbb E\max(A,B)$$

addendum (hint for finding $\mathbb E\max(A,B)$)

$$\int_0^1\int_0^1\max(x,y)dxdy=\int_0^1\int_0^y\max(x,y)dxdy+\int_0^1\int_y^1\max(x,y)dxdy=$$$$\int_0^1\int_0^yydxdy+\int_0^1\int_y^1xdxdy$$

edited Mar 27 '17 at 09:51

answered Mar 27 '17 at 09:23

drhab

151,093

So the expected value of A is the same as the exptected value of max{A, B}? – Andrew C Mar 27 '17 at 09:27
No, it is smaller. See the $<$ sign in the last line of my answer. – drhab Mar 27 '17 at 09:28
But what I can't figure out it what E(max{A,B}) is equal to then? – Andrew C Mar 27 '17 at 09:34
That is something else (as I said) and is not visible yet in your question. So do someting about that. – drhab Mar 27 '17 at 09:36
It is visible now – Andrew C Mar 27 '17 at 09:37
Have you tried to solve it yet by appling $\mathbb E\max(A,B)=\int_0^1\int_0^1\max(x,y)dxdy$? Give that a try. If you get stuck then at least show your efforts on it, and come back. – drhab Mar 27 '17 at 09:40

Calculate $E(max(A, B))$

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