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Let X is random variable with probability density function $f_x(x)$.

If the pdf is scaled by a constant $c$ such that $f_y(y) = c*f_x(x)$, how does the random variable Y corresponding to the pdf $f_y(y)$ relate to random variable X?

Is $f_y(y)$ is a probability density function?

I would like to add more details.

If $f_x(x)\sim\mathcal{X}_v^2$ this also means, $f_x(x)\sim\Gamma(k=v/2, \theta=2)$.

Then does $c*f_x(x)\sim\Gamma(k=v/2, \theta=2c)$?

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    $\begingroup$The pdf must integrate to $1$ across its support so you cannot simply scale it like that as that would integrate to $c$. See en.wikipedia.org/wiki/Location%E2%80%93scale_family$\endgroup$
    – Henry
    CommentedMar 30, 2023 at 9:40
  • $\begingroup$$f_y(y)$ is a probability density function if and only f $c=1$.$\endgroup$CommentedMar 30, 2023 at 9:42
  • $\begingroup$Thanks a lot for the valuable comments. I have added more details to the question. Can you please tell me if I am understanding it correct or not?$\endgroup$
    – amitha
    CommentedMar 30, 2023 at 9:54
  • $\begingroup$Roughly speaking: $Y = cX$ doesn't mean $f_Y(t) = c f_X(t)$$\endgroup$
    – openspace
    CommentedMar 30, 2023 at 10:25

1 Answer 1

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The pdf must integrate to $1$ across its support so you cannot simply scale it like that as that would integrate to $c$. See https://en.wikipedia.org/wiki/Location%E2%80%93scale_family

To take your chi-squared/gamma example of $X \sim \text{Gamma}(k=\frac\nu 2, \theta =2)$ using $k$ as the shape parameter and $\theta$ as the scale parameter, then

  • $Y=cX \sim \text{Gamma}(k=\frac\nu 2, \theta =2c)$ and
  • $Z=\frac{X}{c} \sim \text{Gamma}(k=\frac\nu 2, \theta =\frac2c)$.

Neither has density $c\,f_X(x)$, since that is not a probability density when $c \not=1$. Instead

  • $f_Y(x)=\frac1c f_X(\frac xc)$ and
  • $f_Z(x)= c\,f_X(cx)$
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