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We have two phones and waiting time to get a call are presented by $X$ and $Y$

Let 1st phone to receive on average one call every 2 minutes and second to receive one call on average every 4 minutes. Now we have random variable $Z = min(X;Y)$ What is the $E(Z)$ and $median(Z)$ ?

I approached it that way: $E(Z)= min(E(X);E(Y))$

And according to the problem $Θ(X)=2$ and $Θ(Y)=4$, in Exponential distribution $E(X) = Θ$, so we can conclude that $E(Z) = E(X) = 2$

Is it correct?

And I have no idea how to find the median

Alice
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  • The expectation of the minimum is generally not the minimum of the expectation of the two random variables. See here for a good answer to that problem. –  Sep 28 '17 at 09:04

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The problem with your answer is the step $Z = \min(X,Y) \to E[Z] = \min(E[X], E[Y])$. The expectation is the integral $$ E[Z] = \int \! \min(x,y) \, f_X(x) \, f_Y(y) \, dx \, dy \;, $$ which, as you can see, does not allow you to simply pull the expectation of $X$ and $Y$ into the $\min$ function.

As explained in detailed in the answers to another question, the easiest way to get the distribution of $Z = \min(X, Y)$ is via the cumulative distribution function $F_Z$, which is $$ F_Z(z) = F_X(z) + F_Y(z) - F_{X,Y}(z,z) \;. $$ In case of independent $X$ and $Y$, which I assume is the case here, the joint c.d.f. factors, $$ F_Z(z) = F_X(z) + F_Y(z) - F_X(z) F_Y(z) \;. $$ Here, we have exponential distributions for $X$ and $Y$, so the result is $$ F_Z(z) = 1 - e^{-\frac{3}{4}z} \;. $$ The expectation value is $$ E[z] = \int_{0}^{\infty} \! z \, F_Z'(z) \, dz = \frac{4}{3} \;, $$ or 80 seconds.

The median is that value of $z$ where half the probability mass is below $z$ (and, naturally, half above). In other words $$ F_Z(z) = \frac{1}{2} \implies z = \frac{4 \log(2)}{3} \;, $$ or about 55 seconds.

  • Omg, it is so appealing eventually, honestly, I was confused with this another question explanation, thanks – Alice Sep 28 '17 at 11:10