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Let $X_i$ be i.i.d. $Uniform(5,10)$, and let $Y_t = \sum_{i=1}^t X_i$. Let $T = \mbox{inf}\{t:Y_t \geq 30\}$, what is $\mathbb{E}[T]$?

At first I thought this was similar to Choose a random number between $0$ and $1$ and record its value. Keep doing it until the sum of the numbers exceeds $1$. How many tries do we need?, but after trying some methods mentioned under that question, I still couldn't figure out a solution. I know $T$ can only be $4, 5, 6$, but is it possible to compute each individual probability?

Sean Ian
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1 Answers1

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Let $U_i$ be iid Uniform($0,1$) and let $X_i=5+5U_i$.

Then $$T=\inf\left\{ t\mid U_{1}+\cdots+U_{t}\geq6-t\right\}$$ and: $$\mathbb{E}T=\sum_{n=1}^{\infty}P\left(T\geq n\right)=4+P\left(T\geq5\right)+P\left(T\geq6\right)=$$$$4+P\left(U_{1}+U_{2}+U_{3}+U_{4}<2\right)+P\left(U_{1}+U_{2}+U_{3}+U_{4}+U_{5}<1\right)\tag1$$

With induction we can prove that for $x\in\left[0,1\right]$ we have: $$P\left(\sum_{i=1}^{n}U_{i}\leq x\right)=\frac{x^{n}}{n!}$$ so the last term in $(1)$ equals $\frac1{120}$.

Further the distribution of $U_1+U_2+U_3+U_4$ is symmetric wrt $2$ so that $P\left(U_{1}+U_{2}+U_{3}+U_{4}<2\right)=\frac12$

This together proves that: $$\mathbb ET=4+\frac12+\frac1{120}=\frac{541}{120}$$

drhab
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    You have implicitly said $\mathbb P(T=4)=\frac12$ and $\mathbb P(T=6)=\frac1{120}$ so $\mathbb P(T=5) =\frac{59}{120}$, which would answer "is it possible to compute each individual probability?" – Henry Sep 25 '18 at 15:06
  • @Henry Yes, you are right. Actually I was not aware of that and was only focusing on the mean of $T$. But this is a nice side effect. – drhab Sep 25 '18 at 17:06