Q:

Finding the distribution of $Y$, the minimum of some IID random variables

As I'm learning multivariate distributions and some special distributions such as gamma, chi-sqaure, I'm having a hard time understanding the concepts between the two questions i'm about to write. Following questions are from a textbook from hogg

$Q_1$ : Let $X_1,X_2,X_3,X_4$ be four independent random variables, each with pdf 
  $$f(x) =\begin{cases}
3(1-x)^2, & 0 < x < 1,\\
0,& \text{elsewhere}.
\end{cases}$$
  If $Y$ is the minimum of these four variables, find the cdf and the pdf of $Y$.
  Hint: $P(Y>y) = P(X_i > y, i=1,\dotsc,4)$.
$Q_2$ : Let $X_1,X_2,X_3$ be iid random variables, each with pdf 
  $$f(x) =\begin{cases}
e^{-x},& 0 < x < \infty,\\
0, & \text{elsewhere}.
\end{cases}.$$
  Find the distribution of $Y = \min(X_1,X_2,X_3)$.
  Hint: $P(Y \le y) = 1 - P(Y>y) = 1 - P(X_i > y, i=1,2,3)$

From these two questions above, I don't quite understand how both question is asking for the minimum, yet their hints are totally opposite. Could someone explain this?

A:

In either case, you are asked to find the cdf. Generally, you have the option to find the pdf first or the cdf. Here, you are being guided into finding the cdf first because it happens to be easier.
Now, to find the cdf would be to calculate $P(Y\leq y)$. In words, this means that the minimum $Y$ has to smaller than some constant $y$. But this is a hard approach because you're placing an "upperbound" (for lack if a better word) on the minimum. Instead, it is easier to consider a "lowerbound" on the minimum, $Y>y$. To get this, we notice that
$$P(Y\leq y) = 1- P(Y>y).$$
In words, this means that the minimum has to be larger than $y$. This means that each $X_k$ has to be greater than $y$,
$$P(Y\leq y) = 1- P(Y>y) = 1-P\left(\bigcap_{k = 1}^n X_k >y\right) = 1-\prod_{k=1}^nP(X_k>y)$$
where the last equality is true by independence. Since the $X_k$ are iid, we have
$$1-\prod_{k=1}^nP(X_k>y) = 1 - [P(X_1>y)]^n.$$

