# Calculate distribution of $X$.

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)?

• Do you mean "exponential distribution" whete you put "geometric distribution"? May 28, 2019 at 17:44

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 $$\mathbb E[X_1]=1/\lambda p$$.