If $X$ is a random variable and $a$ is a positive number then does $Y = \max (a,X)$ is a legitimate random variable? If $X$ is random variable with a given PDF and $a$ is a positive number then could $Y = \max (a,X)$ be considered as a legitimate random variable ?

Mathematica result look strange !
For example, when $Y = \max (3,X)$ and $X$ follow the exponential distribution with parameter $\lambda$, could $Y$ be considered as a random variable ?
If so then how can I interpret the result attach in the picture ?
 A: A) Yes, $Y$ is a legitimate random variable as obtained through a legitimate operation (max) between two random variables: $X$ (with its exponential distribution) and a constant random variable equal to $a$ (with its pdf $\delta(x-a)$).
B) Let us work with the cdf of $Y$:
$$F_Y(x)=P(Y<x)=\begin{cases}0&\text{if }x<a\\P(X<x)=F_X(x)=1-e^{\lambda x}&\text{otherwise}\end{cases}\tag{1}$$
You can write (1) under the form:
$$F_Y(x)=e^{-\lambda x}U(x-a)\tag{2}$$
where $U$ is the unit step function at the origin ($0$ for negative arguments, $1$ for positive arguments, with derivative the $\delta(x)$ density function).
If we differentiate the cdf expression (2), we get the pdf into two parts :
$$f_Y(x)=\underbrace{P(X<a)\delta(x-a)}_{\text{accounting for the jump in } x=a (*)}+e^{-\lambda x}$$
(*) : with $P(X<a)=F_Y(a)=(1-e^{-\lambda a})$. Intuitively, by taking the "max", all the "mass" $1-e^{-\lambda a}$ before $x=a$ has been moved  into $x=a$.
It remains a little issue: Mathematica gives for the first term:
$$(1-e^{-\lambda  x})\delta(x-a) \ \text{instead of  } \ \ (1-e^{-\lambda a})\delta(x-a)$$
This is in fact the same thing because, for any function $\varphi$, $\varphi(x)\delta(x-a)$ is the same as $\varphi(a)\delta(x-a)$
(intuition: $\delta(x-a)$ ignores what is outside $x=a$).
Mathematica's result is right!
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Basically $\ds{\pars{~\mbox{besides notation details}~}}$, it's something like
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}\on{P}\pars{x}
\,\delta\pars{y - \max\braces{a,x}}\dd x}
\\[5mm] = &\
\int_{-\infty}^{a}\on{P}\pars{x}
\,\delta\pars{y - a}\dd x +
\int_{a}^{\infty}\on{P}\pars{x}
\,\delta\pars{y - x}\dd x
\\[5mm] = &\
\delta\pars{y - a}\int_{-\infty}^{a}\on{P}\pars{x}
\,\dd x +
\on{P}\pars{y}
\bracks{y > a}
\end{align}

For instance: With $\ds{\on{P}\pars{x} = \bracks{x > 0}\lambda\expo{-\lambda x}}$ and $\ds{\lambda, a > 0}$; the result is given by
\begin{align}
&\delta\pars{y - a}\pars{1 - \expo{-\lambda a}} +
\bracks{y > 0}\lambda\expo{-\lambda y}
\bracks{y > a}
\\[5mm] = &\
\bbx{\delta\pars{y - a}
\pars{1 - \expo{-\lambda a}} +
\bracks{y > a}\lambda\expo{-\lambda y}}
\\ &
\end{align}
