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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 !

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 ?

Tuong Nguyen Minh
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2 Answers2

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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!

Jean Marie
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}
Felix Marin
  • 89,464