The random variable X has homogeneous distrigution I have a problem with exercise:
The random variable X has a homogeneous distribution in the interval [0,1]. Random variable
Y = max(X, 1/2). Please find the expected value of a random variable Y.
I need to do this for discrete variables and also for continuous variable
 A: The formula for expectation of Y:
$$E(Y)=\int_{y\in Y}yf_y(y)dy$$
However, it is easier to think of Y in terms of X:
If $0\leq X\leq.5$, $y$ will be $.5$
If $.5<X\leq1$, $y$ will be $x$.
So we can rewrite the integral as:
$$\int_0^.5.5f_y(y)dx+\int_.5^1xf_y(y)dx$$
Think of it as from $0$ to $.5$, $y$ is always $.5$, so we do not need the variable for $y$, we can just write $.5$, and from $.5$ to $1$, $y$ is always $x$, so we again do not need the variable for $y$. Now, we need to find $f_Y(y)$.
The best way to do this is to find $F_Y(y)$ in terms of x and differentiate with respect to $x$. This is very simple, since:
$F_Y(y)=x$, for all valid $y$, i.e $.5\leq y\leq1$. We differentiate to get:
$f_Y(y)=\frac{d}{dx}x=1$. Since it is $1$, we can just remove it from the intergral, and calculate:
$$\int_0^.5.5dx+\int_.5^1xdx=.625$$
Also, this does not make sense for a discrete variable, as $X$ is homogeneous over $[0,1]$, so it must be continuous. Maybe it was a variation where $X=1,2,...10$, $p_X(x)=.1$ and $Y=$ Max($5,X$)?
