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I’ve managed to solve this problem but the solution I did involves relatively high effort. I think there must be some better alternative I’m overlooking.

A jump has a probability of ${1\over3}$ of being disqualified (and it’s distance is therefore counted as $0$). A qualified jump’s distance is uniformly distributed $\textsf{U}_{[2,3]}$. Find the expectation of the maximum of 3 jumps’ distance.

My solution: Define $$H_i = \begin{cases} 0 & \frac13 \\ \textsf{U}_{[2,3]} & {2 \over 3} \end{cases}. \quad H = \max(H_1,H_2,H_3) $$ Then find $F_{H}(h) = (P(H_i \leq h))^3 $, since the tail formula isn’t very nice we find $f_H(h)$, integrate the expression for the expectation involving it from $2$ to $3$.
If there isn’t one that’s okay, I’m just curious.

kal_elk122
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1 Answers1

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Your approach looks fine.

Alternatively. Let $Y$ be the number of qualified jumps. Then $B$ follows a Binomial distribution with $n=3$ and $p=2/3$. Further,

$$E[H] = E[E[H | Y]] = \sum_{y=0}^3 P(Y=y) E[H|Y=y]$$

But $E[H|Y=y]$ is the expected value of the maximum of $y$ iid random variables uniform on $[2,3]$; which is $\frac{y}{y+1} + 2$.

leonbloy
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  • thank you! can you please explain to me why $E[E[H|Y]] = \sum_{y=0}^3 P(Y=y) \cdot E[H|Y=y]$? – kal_elk122 Oct 12 '22 at 10:22
  • That's the law of total expectation, or "tower rule" https://en.wikipedia.org/wiki/Law_of_total_expectation https://math.stackexchange.com/questions/1148829/understanding-law-of-total-expectation – leonbloy Oct 12 '22 at 11:19