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The time student Anne spent waiting for a school bus has geometric distribution with expectation 10 minutes. Anne can be playful little girl and can sometimes miss a bus so she needs to wait for next. The probability of Anne getting into the bus is $p$. Let $X$ be a random variable representing the time Anne spent at the station until she boarded the bus. Calculate distribution of $X$ and $\mathbb{E}X$.

What is the best method for solving this? Is it by using generating functions (I tried that way but I got stuck)?

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    $\begingroup$ Do you mean "exponential distribution" whete you put "geometric distribution"? $\endgroup$ May 28, 2019 at 17:44

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Let $X_1, X_2, \ldots$ be i.i.d. exponential random variables with $\mathbb E[X_1]=1/\lambda$ and $N\sim\mathrm{Geo}(p)$ be independent of the $X_n$. Then $X=\sum_{i=1}^N X_i$. For each $n\geqslant 1$ the distribution of $X$ conditioned on $N=n$ is given by $$ f_{X\mid N=n}(t) = \frac{(\lambda t)^{n-1}}{(n-1)!}\lambda e^{-\lambda t}, $$ and so the distribution of $X$ is given by \begin{align} f_X(t) &= \sum_{n=1}^\infty f_{X\mid N=n}(t)\mathbb P(N=n)\\ &= \sum_{n=1}^\infty \frac{(\lambda t)^{n-1}}{(n-1)!}\lambda e^{-\lambda t}p(1-p)^{n-1}\\ &= \lambda p e^{-\lambda t}\sum_{n=0}^\infty\frac{(\lambda t(1-p))^n}{n!}\\ &= \lambda p e^{-\lambda t}e^{\lambda t(1-p)}\\ &= \lambda p e^{-\lambda pt}, \end{align} i.e. $X$ has an exponential distribution with mean $\mathbb E[X_1]=1/\lambda p$.

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