Let $X_i$ be iid with uniform distribution on $(0,100)$. Find expected number of $n$ random variables s.t $\sum_{i=1}^{n} X_i \geq 1000$

Question

Let $X_i$ be iid with continuous uniform distribution on $(0,100)$. Find expected number of $N$

Where $$N = \min \{ n : \sum_{i=1}^{n} X_i \geq 1000 \}$$

My thoughts:

$E[\sum X_i] = \sum E[X_i]$, and each $E[X_i] = 50$. So it would take $20$ of $E[X_i]$ to reach $1000$. It seems rather straighfoward. But is this rigorious? Is there a "trap" I missed?

Suppose, instead of $1000$, $N$ was the expected number of trials needed to get $≥50$ . Your logic would tell us that the answer was $1$, but this is clearly false (there is a $\frac 12$ probability that it will take at least $2$). — lulu, Sep 08 '20 at 14:24
@lulu indeed... what would be the right solution to this problem — Chen Chen, Sep 08 '20 at 14:26
Sorry that answer doesn't deal with the case here. See instead the answer below. — StubbornAtom, Sep 08 '20 at 14:43

score 1 · Accepted Answer · answered Sep 08 '20 at 14:27

1

Since $X_i/100 \sim U[0,1]$, $$ \mathsf{E}N=\sum_{k=0}^{10}\frac{(-1)^k}{k!}(10-k)^ke^{10-k}\approx 20.667. $$ (See this question.)

answered Sep 08 '20 at 14:27

Let $X_i$ be iid with uniform distribution on $(0,100)$. Find expected number of $n$ random variables s.t $\sum_{i=1}^{n} X_i \geq 1000$

1 Answers1