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Suppose that we're given four independent random variables $X_1,X_2,X_3$ and $X_4$ and their probability density function is given by:

$f(x)= 3(1-x)^2 $ for $0<x<1$ and otherwise $f(x)=0$.

If $Y$ is the minimum of $X_1,X_2,X_3,X_4$, how can I find the P.D.F and C.D.F of $Y$?

Well, I guess because all random variables are independent we have:

$f(x_1,x_2,x_3,x_4)=81(1-x_1)^2(1-x_2)^2(1-x_3)^2(1-x_4)^2$

I first wanted to say that since $f(x_1,x_2,x_3,x_4)$ is symmetric with respect to $x_1,x_2,x_3,x_4$ I can assume without loss of generality that $0<x_1\leq x_2 \leq x_3 \leq x_4<1$ but then I think it won't help me since it ruins the independence of the random variables.

I don't know what I should do to find the P.D.F of $Y=\min\{X_1,X_2,X_3,X_4\}$. I'm not looking for a full solution at this step, only a hint would be enough. I know how to solve this problem for two variables by considering the cases $X_1 \leq X_2$ and $X_1 > X_2$ separately, but I have no ideas how I can solve this for four variables.

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    $\begingroup$ The probability that $Y \leq t$ is the probability that $X_1 \leq t$ or $X_2 \leq t$ or $X_3 \leq t$ or $X_4 \leq t$. Apply the formula for the probability of the or event and use independence to get the CDF. Now differentiate to get the PDF. $\endgroup$ – Wintermute Jan 12 '14 at 6:59
  • $\begingroup$ @mtiano Sorry but I refuse to do that. See comment on the answer suggesting the same disastrous approach. $\endgroup$ – Did Jan 12 '14 at 8:49
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Hint: express the event that $ Y\ge y $ in terms of the $ X_i $

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Hint 1: $F_Y(y)=\mathbb{P}\{Y\leq y\}=\mathbb{P}\{min\{X_1,X_2,X_3,X_4\}\leq y\}$. Now think about when the event can happen. Clearly it happens if all four RVs are $\leq$ y. It also happens when three are $\leq y$ and one is not. Etc.

Hint 2: Events described above are mutually exclusive. Your$X_i$ are iid sampled from $F_X$

Hint 3: When trying to make sense of Hint 1 think of it in terms of "$n$ choose $k$"

Hint 4: This is straightforward but once you find $F_Y$ you just differentiate and use chain rule to find $f_Y$.

I hope this helps and at the same time doesn't give it all away.

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  • $\begingroup$ To embark on the computation of P(Y<y) rather than P(Y>y) in this context is masochistic. $\endgroup$ – Did Jan 12 '14 at 8:48
  • $\begingroup$ maybe at first, but then you take advantage of being able to write $F_X$ instead of $1-F_X$ every time. I guess you win some you lose some. $\endgroup$ – mathemagician Jan 12 '14 at 9:16
  • $\begingroup$ When considering minimums? Never. $\endgroup$ – Did Jan 12 '14 at 9:20
  • $\begingroup$ you are right, but maybe I wanted OP to get some practice so that he/she can derive the distribution of the j-th ranked RV by him/herself, haha. $\endgroup$ – mathemagician Jan 12 '14 at 9:31
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    $\begingroup$ @wolfies No. To think it is just proves you did not even care to try. $\endgroup$ – Did Jan 17 '14 at 15:36
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The CDF of Y can be written as $$\begin{split} F_Y(y)=&\text{Pr}\left[\min_{k=1,\ldots,4}X_k\leq y\right]\\ =&1-\text{Pr}\left[\min_{k=1,\ldots,4}X_k\geq y\right]\\ =&1-\prod_{k=1}^4\text{Pr}\left[X_k\geq y\right]\\ =&1-\left(1-\text{Pr}\left[X_k\leq y\right]\right)^4\\ =&1-\left(1-F_X(y)\right)^4 \end{split}$$

All you need to find then is $F_X(y)$, and then find the PDF as $\frac{dF_Y(y)}{dy}$

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