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?
$$1-(e^{-\frac{1}{\theta}x})^4$$
Best Answer
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.