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Question Suppose the random variable $N$ denotes the number of jobs in a single-server queue and $X_i$ the service time for the i-th job. Let T denote the time to serve all jobs in the queue, i.e. $T=x_1+x_2+...+x_n$. Prove that $\mathbb{E}(T)=\mathbb{E}(N)\mathbb{E}(X)$.

Attempt to Solution $\mathbb{E}(T)=\mathbb{E}[\mathbb{E}(T|N)]$. Then I derive that $\mathbb{E}(T)=\sum_n Pr(N=n)\sum_i\mathbb{E}(X_i|N)$; however, I failed afterwards by obtaining $\mathbb{E}(T)=\mathbb{E}(X)$, which is obviously wrong.

I wonder where is my mistake and how to derive the correct expectation in this scenario.

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1 Answer 1

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Let $X_1,X_2,\dots $ all have the same probability distribution as $X$.

Then:

$$\mathbb E[T\mid N=n]=\mathbb E[X_1+\cdots+X_n]=n\mathbb EX$$ so that: $$\mathbb E[T\mid N]=N\mathbb EX$$

Then: $$\mathbb ET=\mathbb E[\mathbb E[T\mid N]]=\mathbb E[N\mathbb EX]=\mathbb EN\mathbb EX$$

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