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## Homework Statement

Let Y_1,Y_2 be independent random variable with uniform distribution on the interval [1,2]. Define X=max{Y_1,Y_2}. Find p.d.f., expected value and variance.

## Homework Equations

## The Attempt at a Solution

Since $X=\max\{Y_1,Y_2\}$, this tells $Y_1$ and $Y_2$ must at most $x$. Thus, $\p(\max\{Y_1,Y_2\}\leq x)=\p(\{Y_1\leq x,Y_2\leq x\})=(2-x)^2$ with $1\leq x\leq 2$. So, the $c.d.f.$ of $X$ is

$$\p(x)=

\begin{cases}

0&\text{if $x<1$}\\

(2-x)^2 & \text{if $1\leq x\leq 2$}\\

1& \text{if $x>2$}

\end{cases}$$ where $a,b\in[1,2]$.

Then, the $p.d.f.$ is

$$p(x)=

\begin{cases}

2(2-x)&\;\;\text{if $1\leq a,b\leq 2$}\\

0&\;\;\text{otherwise}

\end{cases}$$

The expected value of $X$ is

$$E(X)=\int_{1}^{2}(4x-2x^2)\,dx=\frac{4}{3}$$

We also have

$$E(X^2)=\int_{1}^{2}(4x^2-2x^3)\,dx=\frac{11}{6}$$

Therefore, the variance is

$$V(X)=E(X^2)-(E(X))^2=\frac{11}{6}-\left(\frac{4}{3}\right)^2=\frac{1}{18}$$