I've got a problem with an exercise. Actually, we consider :
$F(x) = \frac{1}{2}1_{[0;1[}+(1-e^{-x})1_{[1;\infty[}$
We show that it's a cumulative distribution function (of some variable that we will call $X$). Then, we have : $Y=X1_{[0;1[}$ and we want to find the probability distribution of $Y$. So, I wanted to find the cumulative distribution function of $Y$, so I take $x \in \mathbb{R}$ and I want to find :
$F_{Y}(x) = P(Y \leq x) = P(X1_{[0;1[} \leq x) = P(\{w \in \mathbb{R} \, | \, X(w) \leq x\} \cap [0;1[)$
But I just don't find out how to finish this calculus and find the result... It doesn't seem to be the good method cause I don't know $X$...
Anyone could help me ?
Thank you by advance !
I guess what that means is $Y=X1_{[0;1[}(X),$ in other words $Y=X$ if $X\in[0;1[$ and $Y=0$ otherwise.
But looking at the the CDF for $X$, we see that $P(X\le x) = \frac{1}{2}$ for $x\in[0;1[.$ The CDF being flat means there's no probability for $X$ to lie in $]0,1[.$ The fact that the CDF jumps from $0$ to $1/2$ at $X=0,$ means that $P(X=0) = \frac{1}{2}.$ So since $Y=X$ if $X\in[0,1)$ and the only way we have $X\in[0,1)$ is if $X=0,$ we just have $Y=0$ for $X$ in that range. And then we have $Y=0$ otherwise... so the only possibility is $Y=0.$ So the distribution of $Y$ is the degenerate distribution where it is zero with probability one.
Even though it's irrelevant to the distribution of $Y,$ it might be good to finish interpreting the CDF of $X.$ The CDF jumps from $1/2$ to $1-e^{-1}$ at $X=1$ so we have an atom at one with $P(X=1) = 1-e^{-1}-1/2.$ And the CDF is continuous for $x>1,$ indicating that there's density there. Differentiating, the density is $e^{-x},$ so if you, say, wanted to compute $P(X\in [2,3])$ you'd get $$P(X\in [2,3]) = \int_2^3 e^{-x}dx = e^{-2}-e^{-3} $$