Suppose there $\mathbf{2n}$ i.i.d exponential random variables $X_{1},\ldots, X_{n},Y_{1},\ldots,Y_{n}$ with probability density function $$f(x)=\left\{\begin{matrix} e^{-x}, &x\geqslant 0 \\ 0, &x<0 \end{matrix}\right. $$
Now let $Z_{i}=\min(X_{i},Y_{i})$ and $U=\min(X_{*},Y_{*})=\max(Z_{1},\cdots ,Z_{n})$.

So what are the probability density functions of $X_{*}$ and $Y_{*}$?

  • $\begingroup$ What is $X_*$ and $Y_*$ again? $\endgroup$ – Stefan Hansen Feb 7 '13 at 9:05
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    $\begingroup$ While we are at it, what did you try, exactly? It ain't your first question on this exact topic, so it would be good to see that your understanding is getting better... $\endgroup$ – Did Feb 7 '13 at 9:46
  • $\begingroup$ @Did:Thanks for your reply.I had calculated the CDF of Z and U when I posted this question,but I failed to calculate the PDF of X* and Y*. $\endgroup$ – yyzhang Feb 8 '13 at 2:16
  • $\begingroup$ Surely you realize that until now nobody except you knows what X* and Y* are. $\endgroup$ – Did Feb 8 '13 at 6:18
  • $\begingroup$ @Did: All we know is that min(X*,Y*) is the maximum of those $min(X_{i},Y_{i})$.Maybe it is not enough to find the PDF of X* and Y*, but there is no more information about them. $\endgroup$ – yyzhang Feb 8 '13 at 6:34

Computation of the density function of the max or min of a collection of independent random variables is not difficult. They do not even have to be identically distributed, but we stick to that case.

Let $W_1, \dots, W_n$ be iid, and let $S$ be their minimum.

We have $S\gt s$ if and only if all the $W_i$ are $\gt s$. Thus $$\Pr(S\gt s)=\prod\Pr(W_i\gt s).$$ If we know the cumulative distribution functions of $W_i$, you can find the cdf of their minimum, and, if the $W_i$ have continuous distribution, the probability density function of their minimum.

If the $W_i$ all have the same cdf $F_W(w)$, then $S$ has cdf $1-(1-F_W(s))^n$. In the density case, we can differentiate and find that $S$ has density function $nf_W(s)(1-F_W(s))^{n-1}$. (It is best not to try to remember this formula. It is more informative to go through the logic for your concrete case each time.)

Apply this to $Z_i=\max(X_i,Y_i)$. We get $$\Pr(Z_i\gt z)=\Pr(X_i\gt z)\Pr(Y_i\gt z)=e^{-2z}.$$ Thus $Z_i$ has cdf $1-e^{-2z}$ (for $z\ge 0$, and $0$ otherwise).

More generally, the minimum of independent exponentials is exponential, and the parameters add.

Let $T$ be the maximum of $W_1,W_2,\dots,W_n)$. Note that $T\le t$ if and only if all the $W_i$ are $\le t$. Thus $$\Pr(T\le t)=\prod \Pr(W_i\le t).$$ If you know the cdf's of the $W_i$, you can use the above fact to find the cdf of $T$, and then in the continuous case the density function of $T$.

In the particular case where all the $W_i$ have the same cdf $F_W(w)$, the cdf of $T$ is $(F_W(t))^n$. In the density case, it follows that the probability density function of $T$ is $nf_W(t) (F_W(t))^{n-1}$.

Now you have all the material you need to find the distribution of the max of your $Z_i$, and some related quantities.

Added: So explicitly for $\max(Z_1,\dots,Z_n)$ the density function is $2ne^{-2t}(1-e^{-2t})^{n-1}$ (for t \ge 0, and $0$ elsewhere).

  • $\begingroup$ Thank you very much. I agree with what you said, but question is how to find the probability density functions of X* and Y*. Can you give me any idea? $\endgroup$ – yyzhang Feb 8 '13 at 1:46
  • $\begingroup$ Tomorrow (it is late) I can do detail, if necessary (it shouldn't be!), the density of $\max(Z_1,\dots,Z_n)$. It is not possible for me to find density functions of $X_\ast$ and $Y_\ast$, since nowhere in your post are these defined. $\endgroup$ – André Nicolas Feb 8 '13 at 6:01
  • $\begingroup$ I have calculated the density of $U=max(Z_{1},\cdots ,Z_{n})$. min(X*,Y*),identical to U, is the maximum of these $min(X_{i},Y_{i})$.This is all what we know and I am not sure if it is possible to find the density functions of X* and Y*. $\endgroup$ – yyzhang Feb 8 '13 at 6:23
  • $\begingroup$ You have changed the notation from $X_\ast$ to $X^\ast$. You still have not said what $X^\ast$ and $Y^\ast$ mean in your course. I could guess that $X^\ast$ is the minimum of the $X_i$. I could guess that it is the maximum of the $X_i$. I could guess something else, like the mean. $\endgroup$ – André Nicolas Feb 8 '13 at 6:34
  • $\begingroup$ :Sorry, but these two notations are the same.I fails to type the former one today. As I said, all we know is that min(X*,Y*) is the largest one of those $min(X_{i},Y_{i})$. Of course, we can know the distribution of it. But in my question, I want to know the PDF of X* and Y* in min(X*,Y*). Just like Did said, maybe the information is not enough to find these PDF, but there is no more information. $\endgroup$ – yyzhang Feb 8 '13 at 6:53

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