Question

Let X have probability density f X. Find the probability density function of the random variable Y defined by Y = a X + b.

Short answer

We have to consider all three cases $a>0,a<0,a=0$

Step-by-step solution

Step:1 Given Information

Assume that X has a probability density of f X. Find the probability density function for the random variable Y, which has the formula

$Y=aX+b$.

Step:2 Definition

A probability density function (PDF) is used in probability theory to signify the random variable's likelihood of falling into a particular range of values as opposed to taking up a single value. The feature illustrates the normal distribution's probability density function and how mean and deviation are calculated.

Step:3 Explanation of the solution

We have three cases. Suppose that $a>0$. The CDF of Y in this case is

$F_Y\left(y\right)=P(Y\le y)=P(aX+b\le y)=P\left(X\le \frac{1}{a}(y-b)\right)=F_X\left(\frac{1}{a}(y-b)\right)$

Use differentiation to obtain the PDF.

$f_Y\left(y\right)=\frac{d{F}_y}{dy}\left(y\right)=f_X\left(\frac{1}{a}(y-b)\right)·\frac{1}{a}$

Now, suppose that $a<0$. The CDF of Y in this case is

$F_Y\left(y\right)=P(Y\le y)=P(aX+b\le y)=P\left(X\ge \frac{1}{a}(y-b)\right)=1-F_X\left(\frac{1}{a}(y-b)\right)$

Use differentiation to obtain the PDF.

$f_Y\left(y\right)=\frac{d{F}_y}{dy}\left(y\right)=-f_X\left(\frac{1}{a}(y-b)\right)·\frac{1}{a}$

If $a=0$, we have that $Y=b$, so Y is equal to b almost certainly.