Finding maximum of two variables Given $X$ is uniform on $[0, 10]$. Let $$Y = \max(5, X).$$ Determine Var(Y).
I'm familiar with how to find the variance of a uniform random variable, as well as the max of two random variables. However, I'm a bit confused about the 5.
For the max order statistic:
$P(\text{max}(5,X) \leq Y) = P(X \leq y)P(5 \leq y)$
Beyond this I'm stuck as to what to do next. 
Also, I know that the variance would be $\frac{(b-a)^2}{12}$ for a uniform random variable on interval $[a,b]$.
 A: Hint:
$Y$ and $Y^2$ are nonnegative random variables so that:
$$\mathbb{E}Y=\int_{0}^{\infty}P\left(Y>y\right)dy=\int_{0}^{5}1dy+\int_{5}^{10}P\left(X>y\right)dy$$
and:
$$\mathbb{E}Y^{2}=\int_{0}^{\infty}P\left(Y^{2}>y\right)dy=\int_{0}^{25}1dy+\int_{25}^{100}P\left(X^{2}>y\right)dy$$
A: The cdf of $Y$ is
$$P(Y\leq y)=P(\max(5,X) \leq y) =$$
$$=P(\max(5,X) \leq y\cap X\leq5)+P(\max(5,X) \leq y\cap X>5)=$$
$$=P(X\leq 5\leq y)+P(5<X\leq y)$$
$$=
\begin{cases}
\frac12+\frac{y-5}{10},& \text{ if } 10\ge y>5\\
0,&\text{ if } y\le 5\\
1,&\text{ if } y>10. 
\end{cases}$$
So, $Y=5$ with probability $\frac12$ and is uniformly distributed over $[5,10]$ also with probability $\frac12$.
Then
$$E[Y]=\frac12 5 +\frac12 \frac{15}2=\frac{25}4.$$
I just hope that calculating the variance will not cause any problems.
A: $E(Y)=\int\limits_{0}^{10} \frac{max(X,5)}{10}dx=\int\limits_{0}^{5}\frac{5}{10}dx+\int\limits_{5}^{10}\frac{x}{10}dx$, as if $0 \leq X \leq 5$ then $max(X,5)=5$and if $5 \leq X \leq 10$, then $max(X,5)=X$.
