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The question is as follows:

You arrive at the post office and as you enter, each of the two clerks, Jim and Jack, starts serving a client. The amount of time needed by Jim to serve his client is X minutes, and the amount of time needed by Jack to serve his client is Y minutes, where X and Y are independent exponential random variables, X having parameter (rate) 1/2 per minute and Y having parameter 1/4 per minute.

Q) Since you are next in line, you will be served by whichever clerk is first to finish serving his current client. Find the probability that you will be served by Jim.

So Jim is represented by X with rate = 1/2
Jack is represented by Y with rate 1/4

The question is asking for P ( X$\lt Y) $

a) I don't understand where the inner part of the integral $P(X\lt Y | X=x)P(X=x)$ is coming from.

b) How can you just replace P$(x\lt Y)$ with $e^{\frac{-x}{4}}$ ?

c) The p.d.f is $\frac{1}{4}e^{\frac{-x}{4}}$ where did the $\frac{1}{4}$ go?

The given solution are as follows:

enter image description here

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1 Answer 1

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a) This is an application of the law of total probability (or equivalently the law of total expectation applied to an indicator random variable). You can probably find it in the section of your textbook or lecture notes that discusses joint continuous distributions.

b) If $Y$ is an exponential random variable with parameter $\lambda$, then $P(Y>x) = e^{-\lambda x}$. (If you remember the PDF, then you can re-derive this from scratch by $P(Y>x) = \int_x^\infty \lambda e^{-\lambda t} \, dt = [e^{-\lambda t}]_{t=x}^\infty = e^{-\lambda x}$.)

c) You ended up not needing the PDF of $Y$ in those computations.

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  • $\begingroup$ how is P(X<Y|X=x) replaced by P ( x<Y) from first to second line? is it using P(A|B) = P(A & B) / P(B) , then replaced P(A&B) = P(A)P(B) ?, how can it just assume independence to do that? $\endgroup$
    – Kal
    Apr 21, 2019 at 21:52
  • $\begingroup$ Yes $\mathsf P(X{<}Y)=\int_\Bbb R \mathsf p(X{<}Y, X{=}x)~\mathrm d x$. Then since the independence is stated in the problem, we have: $$\mathsf P(X{<}Y)~{=\int_\Bbb R \mathsf P(x{<}Y\mid X{=}x)~\mathsf p(X{=}x)~\mathrm d x\\=\int_\Bbb R \mathsf P(x{<}Y)~\mathsf p(X{=}x)~\mathrm d x}$$ $\endgroup$ Apr 22, 2019 at 11:53

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