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I'm working on this problem:

Let $X_1, X_2, X_3,$ and $X_4$ be independent random variables from the exponential distribution with mean $\theta$. Let $Y =$min$(X_1,X_2,X_3,X_4)$

(a) Find the cdf of Y
(b) Find the pdf of Y. What is the distribution of Y?

So I know that the cdf of an exponential function with a single random variable would look like this:

$$1-e^{-\frac{1}{\theta}x}$$

And a similar problem but with a uniform distribution you'd end up multiplying the cdf's and essentially end up with the cdf to a power of 4.

So does the same logic apply? Could I multiply the cdfs for these random variables to get an answer like this?

$$1-(e^{-\frac{1}{\theta}x})^4$$

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You are almost right.

$P(Y\leq y)=1-P(min(X_1,X_2,X_3,X_4)\geq y)$

$=1-P(X_1\geq y,X_2\geq y,X_3\geq y,X_4\geq y)$

$X_i$ are independent variables. Thus

$1-P(X_1\geq y)\cdot P(X_2\geq y)\cdot P(X_3\geq y)\cdot P(X_4\geq y)$

$1-e^{-\frac1{\theta} y}\cdot e^{-\frac1{\theta} y}\cdot e^{-\frac1{\theta} y}e^{-\frac1{\theta} y}\cdot $

$P(Y\leq y)=1-e^{-\frac4{\theta} y}$

Thus the cdf is

$$F(y;\theta)=\begin{cases} 1-e^{- \frac4{\theta} y}, y\geq 0\\0, \quad y <0 \end{cases}$$

The pdf can be evaluated by taking the derivative.

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  • $\begingroup$ Great answer! Thanks for the help $\endgroup$ – kojak Jan 19 '17 at 19:24
  • $\begingroup$ You are welcome. You wasn´t far away from the right solution. $\endgroup$ – callculus Jan 19 '17 at 19:26

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