Distribution of Y = max(X, alpha) I have a random variable $X$ with density $f_X(x) = \dfrac{1}{(1+x)^2}, x > 0$. I have a transformation $Y=\max(X, \alpha)$ where $\alpha > 0$ and it is a constant. I need to find the distribution of Y.
I've tried to proceed using  the CDF of Y.
$$\mathbb{P}(Y \leq y) = \Bbb P \Bigl(\max\bigl(X, \alpha\bigr) \leq y\Bigr)$$
I rewrite the last event like this
$$\mathbb{P}\Bigl(\max\bigl(X, c\bigr) \leq y\Bigr) = \mathbb{P}\bigl(X \geq y\bigr)\cdot\mathbb{P}\bigl(\alpha \geq y\bigr)$$
but for me $\mathbb{P}(\alpha \geq y)$ does not make any sense.
What can I do to find the distribution of $Y$?
 A: a simple calculation shows that
\begin{align}
\Bbb P \Bigl(\max\bigl(X, \alpha \bigr) \leq y  \Bigr)& =  \\[2mm]
\Bbb P \Bigl(\max\bigl(X, \alpha \bigr) \leq y, \ X < \alpha  \Bigr)& +
\Bbb P \Bigl(\max\bigl(X, \alpha \bigr) \leq y, \ X \geq \alpha  \Bigr) = \\[2mm]
\Bbb P \Bigl(\alpha \leq y, \ X < \alpha  \Bigr)& +
\Bbb P \Bigl(X \leq y, \ X \geq \alpha  \Bigr) = \\[2mm]
\Bbb P \Bigl(\alpha \leq y, \ X < \alpha  \Bigr)& +
\Bbb P \Bigl(\alpha \leq X \leq y \Bigr) .
\end{align}
Now, if $\alpha >y$ then both terms in the last line are equal to zero. So we only need to further investigate the case where $\alpha \leq y$. If this is the case then we get using the continuity of the distribution of $X$
\begin{align}
\Bbb P \Bigl(\alpha \leq y, \ X < \alpha  \Bigr)& +
\Bbb P \Bigl(\alpha \leq X \leq y \Bigr) = F_X(\alpha) + F_X(y) - F_X(\alpha) = F_X(y).
\end{align}
To sum up
$$
F_Y(y) = \Bbb P (Y \leq y) = \begin{cases} F_X(y) =  \Bbb P (X \leq y), & \text{if} \ y < \alpha,\\
 0 , & \text{if} \ y \geq \alpha.\end{cases} 
$$
