Find PDF of the random variable Y. Please help!

Question

Let $X_1$ and $X_2$ be independent and identically distributed Uniform $(0,1)$ random variables. Let $Y = \max(X_1, X_2)$. Find the PDF of the random variable $Y$.

I am having a hard time progressing with this problem. Would I first begin with $Fy = P(\max(X_1, X_2) \le z)$, and then take the derivative?

Answer 1

If $Y \in (0, b)$ (for $b \in [0, 1]$), then we know that both $X_1 \in (0, b)$ and $X_2 \in (0, b)$. Then, $$\begin{align*} \mathrm{Pr}(Y \leq b) &= \mathrm{Pr}(X_1 \leq b \cap X_2 \leq b) \\ &= \mathrm{Pr}(X_1 \leq b) \cdot \mathrm{Pr}(X_2 \leq b) \tag{by independence} \\ &= (b)(b) \\ \mathrm{Pr}(Y \leq b) &= b^2. \end{align*}$$ Then we know that the CDF of $Y$ is $b \mapsto b^2$.

Let $f$ be the PDF of $Y$. Then $$\begin{align*} \mathrm{Pr}(Y \leq b) &= \int_{0}^{b} f(y) \, dy \\ b^2 &= \int_{0}^{b} f(y) \, dy \\ f(y) &= 2y. \tag*{$\square$} \end{align*}$$

Answer 2

$P(Y < t) = P (\max(X_1, X_2) < t) = P (X_1 < t ~\wedge X_2 < t) = P(X_1<t)P(X_2<t) $ by independence. Calculate this cumulative distribution function and it should be easy to find the density

Answer 3

Note that \begin{align} Y:\Omega&\longrightarrow \mathbb R\\ \omega&\longmapsto \max\{X_1(\omega),X_2(\omega)\} \end{align} Then $F_Y(y):=\mathbb P(Y\leq y) =\mathbb P(\max\{X_1,X_2\}\leq y) =\mathbb P(X_1\leq y, X_2 \leq y)$. If they are independent then $$F_Y(y)=\mathbb P(X_1\leq y)·\mathbb P(X_2\leq y) = F_{X_1}(y)·F_{X_2}(y).$$