This is a pretty basic question, I understand, but in my head there is something that doesn't make sense.

Let's say I have a uniform probability density function X that goes from [0,6]. The pdf would thus be 1/(6-0), which is just 1/6.

Now, let's say I multiply this pdf X by 2, to get Y= 2X. Would the pdf then be 2*(1/6)? If so, wouldn't this break the condition that the integral of a pdf has to be equal to 1? What would I do? And for whatever answer I get, would that work for all pdf's (binomial, Poisson, uniform, normal, etc).

  • 1
    $\begingroup$ Why should scaling a pdf give another pdf? $\endgroup$
    – Randall
    Feb 16, 2021 at 18:10
  • $\begingroup$ If $Y=2X$ then $Y$ takes values on $[0,12].$ And the pdf of $Y$ is $1/2$ of the pdf of $X:$ that is $1/12.$ $\endgroup$
    – mfl
    Feb 16, 2021 at 18:16
  • 2
    $\begingroup$ No... You seem to be confusing random variables with probability density functions. A random variable $X$ who is uniform over $[0,6]$ you can talk about multiplying by $2$ to get a random variable $Y=2X$ who is uniform over $[0,12]$. You may not multiply a probability density function by a scalar to get another probability density function... again for the very reason you cite... the integral of a pdf must equal $1$. If you were to have scaled a pdf by $k$ the integral would no longer be $1$ but would instead be $k$. $\endgroup$
    – JMoravitz
    Feb 16, 2021 at 18:19
  • $\begingroup$ As for "...would be just $\frac{1}{6}$..." No. The probability density function for the uniformly distributed random variable over $[0,6]$ would be $f(x)=\begin{cases}\frac{1}{6}&\text{if }0\leq x\leq 6\\ 0&\text{otherwise}\end{cases}$. The second half of that is crucial and can not be left off. $\endgroup$
    – JMoravitz
    Feb 16, 2021 at 18:22

2 Answers 2


You can explain it in this way: $$X \sim Uni(0,6) \Rightarrow F_X(t)=P(X\leq t) = \frac{t}{6} \\ Y = 2X \Rightarrow F_Y(s) = P(Y\leq s) = P(2X \leq s ) = P (X\leq \frac {s}{2}) =_{s\in(0,12)} \frac{s}{12}$$

Hence it is similar to what you thought, but opposite, we got that the density function of $Y$ is $\frac{1}{12}$ which is half, and the explanation of this is in the computation above

The same explanation goes for every other distribution, BUT that doesn't mean you will get the property of anti-homogeneous we got here.

For example, if $X ~ \exp(1)$ then $f_X(x)=e^{-x}$ and for $Y=2X$ we get $$ F_Y(t) = P(Y\leq t) = P(X \leq \frac{t}{2}) = 1-e^{-\frac{t}{2}}$$ Hence, in this case $f_Y(y) = e^{-\frac{y}{2}}$ which means $Y\sim\exp(\frac{1}{2})$ and NOT $(\frac{1}{2}\exp{1})$ as you could (maybe) have expected


Using fundamental transformation theorem you get

$$f_Y(y)=f_X\left(g^{-1}(y)\right)|\frac{d}{dy} g^{-1}(y)| $$



$$y=2x\rightarrow x=\frac{y}{2}\rightarrow x'=1/2$$

In other word, $Y\sim U(0;12)$

Important observation: in your case you are not multiplying the distribution by 2 but THE RANDOM VARIABLE. It is not the same.


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