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:
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?