Cumulative distribution function of $X1_{X>c}$ 
I want to find the CDF of $Y=X1_{X>c}$. Using some of my knowledge I have found it to be $$P(Y<y)=P(X>c)P(X<y)+P(X<c)$$ where $c$ is some positive constant. The random variable $X$ is defined only for positive values. I want to know whether my expression for $P(Y<y)$ is right or wrong? Please correct me if my expression is wrong.

My Attempt:
\begin{align}&P(Y<y)=P(X1_{X>c}<y)=\\[0.2cm]&=P(\mathbf 1_{X>c}=1)P\left(X<\frac{y}{\mathbf 1_{X>c}}\middle |\: \mathbf 1_{X>c}=1\right)+P(\mathbf 1_{X>c}=0)P\left(X<\frac{y}{\mathbf 1_{X>c}}\middle |\: \mathbf 1_{X>c}=0\right) \\[0.2cm]&=P(c<X<y)+P\left(\mathbf 1_{X>c}=0\right)P(X<\infty)\end{align}
 A: The CDF would be
$$P(Y \leq y)=P(Y \leq y \mid X \leq c)P(X \leq c)+P(Y \leq y \mid X>c)P(X>c).$$
Now you're left to evaluate the conditional probabilities:


*

*For $P(Y \leq y \mid X>c)$, we notice that when $X>c$ we have $Y=X$. Thus $P(Y \leq y \mid X>c)=P(X \leq y \mid X>c)=\frac{P(c<X \leq y)}{P(X>c)}$. This will be $0$ if $y \leq c$, otherwise it will be $\frac{P(X \leq y)-P(X \leq c)}{P(X>c)}$.

*If $X \leq c$ then $Y=0$, so $P(Y \leq y \mid X \leq c)=1$ if $y \geq 0$ and $0$ otherwise.


You can now substitute in to finish the problem. 
A: First write down $Y$ explicitly $$Y=X\mathbf 1_{X>c}=\begin{cases}0, &\text{ if }X\le c\\X, &\text{ if } X>c\end{cases}$$
So, the value of $Y$ depends on the value of $X$ and hence when calculating a probability about $Y$ you should take cases based on the value of $X$. For the CDF (which I define with $\le$ instead of $<$ as in your post) we need only to consider values of $y>0$ since $X>0$ and hence $F_Y(y)=0$ for $y\le 0.$ So, for $y>0$ you have 
\begin{align}F_Y(y)&=P(Y\le y)=P(X\mathbf 1_{X>c}\le y)\\[0.3cm]&=P(0\le y\mid X\le c)P(X\le c)+P(X\le y\mid X>c)P(X>c)\\[0.2cm]&=P(0\le y)P(X\le c)+\frac{P(c<X\le y)}{P(X>c)}P(X>c)\\[0.2cm]&=1\cdot P(X\le c)+P(c<X\le y)\\[0.2cm]&=P(X\le c)+P(c<X\le y)\end{align}
Now, the second summand is equal to zero if $0<y\le c$ which gives $$F_Y(y)=P(X\le c)+0, \qquad 0<y\le c$$ Contrary, if $y>c$, then you may add the probabilites to get $$F_Y(y)=P(X\le c)+P(c<X\le y)=P(X\le y), \qquad y>c$$ Putting these together, you get \begin{align}F_Y(y)=\begin{cases}0,&\text{if }y\le 0 \\F_X(c),& \text{if }0<y\le c \\F_X(y),& \text{if }c<y\end{cases}\end{align} This shows that $Y$ has an atom at $y=0$, i.e. $P(Y=0)>0$.
