0
$\begingroup$

This is a question from a previous exam. It seems too easy for the high amount of points it was worth. I don't have the answer key. Critique my answers please?

$X$ and $Y$ are independent uniform random variables with ranges (0,1) and (0,1). Define $W$ as the maximum of ($X$, $Y$): $W$=Max($X$,$Y$).

  1. Compute $P(W\leq0.3)$

$P(W\leq0.3) = P(X\leq0.3)\cdot P(Y\leq0.3) = 0.09$

  1. Compute $P(W=X)$

$0.5$ by symmetry

  1. Compute $E[W]$

$\int_0^12w^2dw = \frac{2}{3}$

  1. Compute $E[W|W=X]$

$\frac{2}{3}$ by symmetry and independence. I interpreted $W=X$ to mean the realized value of W is equal to the realized value of X. Is that the correct interpretation?

$\endgroup$
1
  • 1
    $\begingroup$ You can get nicer spacing for conditional probabilities by using \mid instead of |. $\endgroup$
    – joriki
    Aug 11 '18 at 17:13
1
$\begingroup$

Your answers are all right, though your reasoning is a bit terse. For $E[W]$, it's not clear to me what you computed – I guess you took the cumulative distribution function from $1.$, differentiated it and integrated it with $w$? I don't see where you need independence in $4.$; I think the result follows from $3.$ simply by symmetry. Here and in $2.$, to be precise you need not only symmetry but also $P(X=Y)=0$. I think your interpretation of the notation $E[W\mid W=X]$ is correct.

$\endgroup$

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.