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Let $X$ be a continuous random variable with density function $f_X$ and let $a,b>0$.

What is $Y=aX+b$?

I need some help with this one. And I am quite sure it is not $af_X+b$.

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    $\begingroup$ The literal answer to the question asked: "What is $Y = aX+b$?" is that $Y$ is a random variable that happens to be a function of the random variable $X$. $\endgroup$ Commented Jun 2, 2015 at 12:58

2 Answers 2

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We have that $$ \Pr\{aX+b\le y\}=\Pr\{X\le(y-b)/a\}=\int_{-\infty}^{(y-b)/a}f_X(x)\mathrm dx. $$ Using the substitution $x=(t-b)/a$, we obtain that $$ \Pr\{Y\le y\}=\int_{-\infty}^{(y-b)/a}f_X(x)\mathrm dx=\int_{-\infty}^y\frac1af_X((t-b)/a)\mathrm dt $$ and the density function $f_Y(y)=a^{-1}f_X((y-a)/b)$.

In general, if $Y=g(X)$ with a monotone function $g$, we have that $$ f_Y(y) = \left| \frac{\mathrm d}{\mathrm dy} (g^{-1}(y)) \right| \cdot f_X(g^{-1}(y)), $$ where $g^{-1}$ denotes the inverse function (see here for more details). In this particular case $g(x)=ax+b$ for $x\in\mathbb R$.

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  • $\begingroup$ Killing flies with a sledge-hammer? And not an answer to the question asked: "What is $Y = aX+b$?" $\endgroup$ Commented Jun 2, 2015 at 12:56
  • $\begingroup$ Where you use the fact that $a,b>0$? $\endgroup$
    – Meet Patel
    Commented Sep 29, 2023 at 3:14
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    $\begingroup$ @MeetPatel The sign of $b$ does not affect the derivation but the sign of $a$ does. The assumption that $a>0$ is used when we divide both sides of the inequality $aX+b\le y$ by $a$ (we do not reverse the inequality). $\endgroup$
    – Cm7F7Bb
    Commented Sep 29, 2023 at 10:56
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$X$ is a random variable, wich means $X$ is a measurable function $X:\Omega\rightarrow\mathbb R$ where $\mathbb R$ is equipped with the Borel-$\sigma$-algebra. In that context $Y:\Omega\rightarrow\mathbb R$ prescribed by $\omega\mapsto aX(\omega)+b$ is also a random variable. Here I presume that $(\Omega,\mathcal A,P)$ is the underlying probability space.

For every Borelset $C$:$$P\left\{ Y\in C\right\} =P\left\{ aX+b\in C\right\} =\int1_{C}\left(ax+b\right)f_{X}\left(x\right)dx$$

Substitution $y=ax+b$ shows that the RHS equals:

$$\int1_{C}\left(y\right)\frac{1}{a}f_{X}\left(\frac{y-b}{a}\right)dy=\int_{C}\frac{1}{a}f_{X}\left(\frac{y-b}{a}\right)dy$$

This proves that the function prescribed by: $$y\mapsto\frac{1}{a}f_{X}\left(\frac{y-b}{a}\right)$$ serves as PDF of $Y$.

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  • $\begingroup$ OK, I missed the $a > 0$ part. But I still think it is overkill. $\endgroup$ Commented Jun 2, 2015 at 13:47
  • $\begingroup$ $$ P\left\{ aX+b\in C\right\} =\int1_{C}\left(ax+b\right)f_{X}\left(x\right)dx$$ .Can you explain this equality? i know about For every Borelset $C$:$$P\left\{ X\in C\right\} =\int_{C} f_{X}\left(x\right)dx$$ . $\endgroup$
    – Meet Patel
    Commented Sep 28, 2023 at 18:09
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    $\begingroup$ @MeetPatel What I used is the general equality $\mathbb Eh(X)=\int h(x)dF_X(x)=\int h(x)f_X(x)dx$. This for function $h$ prescribed by $x\mapsto 1_C(ax+b)$. Not familiar with it? Back to the basics then! Expectations are not more than integrals and probability theory is filled up with them. If a CDF has a derivative then this derivative serves as density, yes. But "CDF having a derivative" is not necessary for the existence of a density. $Y$ can have a density while at the same time its CDF is not differentiable. Anyway, I will stop here because this is a place for comments (not college). $\endgroup$
    – drhab
    Commented Sep 29, 2023 at 10:11
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    $\begingroup$ @MeetPatel I think that is correct but not more than that. Just like you I must first check it (e.g. on the internet). Just not in the mood for that. Further it has no connection with the answer we are both commenting. $\endgroup$
    – drhab
    Commented Sep 29, 2023 at 11:22
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    $\begingroup$ @drhab Thank you for your response . $\endgroup$
    – Meet Patel
    Commented Sep 29, 2023 at 11:24

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