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If I am given $X$ that follows an exponential distribution with mean $m$ and $Y$ that follows a poisson distribution with mean $n$, how can I use them to find the conditional probability density function of$ X$ given $Y=y$? $X$ and $Y$ are dependent.

The waiting time for a customer to be served at a bank follows an exponential distribution with mean $5$. On any particular day, the number of customers visiting the bank follows a Poisson distribution with mean $10$. Define the total waiting time $S_N = X_1 + X_2 + · · · + X_N$ where $X_i$ is the waiting time of the $i$th customer, and $N$ is the number of customers.

This is the question I'm given to find the conditional probability density function of $S_N$ given $N=n$.

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    $\begingroup$ The question in the first paragraph does not make sense unless one is given the relationship between X and Y, that is, the joint distribution of (X,Y). $\endgroup$ – Did Mar 27 '13 at 15:47
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Once you have decided to condition on $N=n$, it doesn't matter what the distribution of $N$ is. $N=n$ means you had $n$ customers, and for each customer the service time was a (presumably independent) $\mathrm{Exp(5)}$ distributed random variable. So you have to work out the distribution of the sum of $n$ $\mathrm{Exp(5)}$ RVs.

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  • $\begingroup$ which is presumably the Erlang distribution $\mathrm{Erlang(n,5)}$. $\endgroup$ – Caran-d'Ache Sep 27 '13 at 10:21

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