Suppose that we have a random variable X with probability distribution $PDF_X(\mu_x,\sigma_x)$

Consider random variable $Y=\frac {X-a}b$ , I know that mean and variance of Y would be:

$$\mu_y=\frac {\mu_x-a}b, \sigma_y=\sqrt{ \frac {\sigma_x^2} {b^2} }$$

Would X and Y have the same type of probability distribution (Of course with different mean and variance)?

For example I know that if X is a Normal random variable, Y would be again a Normal random variable. Is this true for all the other probability distribution?

Thank you.

  • 1
    $\begingroup$ That depends on our definition of 'same'. No information is lost in the mapping (that is, $I(X;Y)=H(X)$), since the linear map is invertible. However, some properties of the distribution may clearly change, such as positivity for example. $\endgroup$ Sep 22, 2013 at 15:41

2 Answers 2


Informally, if we have a random variable $X$, and $Y=aX+b$, where $a$ and $b$ are constants and $b\ne 0$, then $X$ and $Y$ are close relatives.

But the distributions of $X$ and $Y$ need not have the same type of name. As you pointed out, if $X$ has normal distribution, then so does $Y$. Similarly, if $X$ has uniform distribution, so does $Y$.

However, if $X$ has binomial distribution, then $aX+b$ only has binomial distribution if $a=1$ and $b=0$. Similar comments could be made about the hypergeometric, the Poisson, and many others.

The fact that if $X$ has binomial distribution, then (usually) $aX+b$ does not has no real mathematical significance. It just has to do with the kind of distributions we choose to call binomial. The very close relationship between $X$ and $aX+b$ remains, even if we do not happen to give their distributions the same name.

  • $\begingroup$ In general, how can I find the PDF of Y? $\endgroup$
    – Alex
    Sep 22, 2013 at 17:37
  • $\begingroup$ Is it $F_Y(y)=aF_X(\frac{y-b}a)+b$? $\endgroup$
    – Alex
    Sep 22, 2013 at 17:39
  • 3
    $\begingroup$ In the discrete case, we have $\Pr(Y=y)=\Pr(aX+b=y)=\Pr(X=\frac{y-b}{a})$. Now use the known facts about $X$. In the continuous case, the density function of $Y$ is $\frac{1}{|a|}f_X((y-b)/a)$, where $f_X$ is the density function of $X$. $\endgroup$ Sep 22, 2013 at 17:47
  • $\begingroup$ thanks. but why the continuous case is different than discrete case? and why the continuous case is like that? $\endgroup$
    – Alex
    Sep 22, 2013 at 19:17
  • $\begingroup$ At the cumulatve distribution function level, the answer for the continuous case is the same as for the discrete. Density is different. For that, in the continuous case, we differentiate the cdf. That's how I got the density function of the previous comment. $\endgroup$ Sep 22, 2013 at 19:30

$c\cdot X$ does not necessarily follows the same family of distributions as $X$; depends on the family of distribution $X$ belongs to (Not true e.g. for the Beta distribution, but true for Normal).

  • $\begingroup$ Welcome to the site. You should use mathjax for formatting. This time I've done it for you $\endgroup$
    – Jakobian
    Sep 26, 2018 at 18:28

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