Suppose we specify the gamma pdf in the following format:

$f(x) = \lambda e^{-\lambda x} \frac{(\lambda x)^{n - 1}}{(n - 1)!}$

Further suppose we want the distribution of a $\text{gamma}(\lambda = 1,n = 2)$ random variable conditional on its value exceeding 5.

Now we can say that the pdf of this random variable is defined as:

$f(x) = \frac{xe^{-x}}{\int_{5}^{\infty} x e^{-x}dx}$

Why is the above-mentioned true?

I know that:

$f_{X|Y} = \frac{f(x,y)}{f(y)}$

But how would the above-mentioned equality be applied in my case? It is easy to see that the denominator in $\frac{xe^{-x}}{\int_{0}^{\infty} x e^{-x}dx}$ is defined as the probability that x is greater than 5. Which conforms to the usual definition of $f_{X|Y}$. How , if even, was the joint probability density function calculated in this case?

  • $\begingroup$ You are using a joint density approach when $Y=\mathbb I_{X>5}$ does not have a density proper. $\endgroup$
    – Xi'an
    Commented Nov 15, 2021 at 20:48
  • $\begingroup$ Not sure what you mean by that. $\endgroup$ Commented Nov 15, 2021 at 20:51
  • $\begingroup$ Why is $xe^{-x}$ in the numerator? $\endgroup$ Commented Nov 15, 2021 at 20:56
  • 1
    $\begingroup$ This question is asked and generally answered at stats.stackexchange.com/questions/525894. The keyword to use in a search is "truncated distribution." $\endgroup$
    – whuber
    Commented Nov 15, 2021 at 21:14
  • $\begingroup$ BTW, neither expression you give for the two versions of $f$ is fully correct: the first needs to indicate $f$ is zero for $x\lt 0$ and the second needs to indicate that $f$ is zero whenever $x\lt 5.$ Perhaps this resolves some of your questions? $\endgroup$
    – whuber
    Commented Nov 15, 2021 at 21:31

1 Answer 1


In the definition of the conditional density $$f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)}$$ both $f_{X,Y}(\cdot,\cdot)$ and $f_Y(\cdot)$ are densities wrt some appropriate dominating measures. You need to find the proper dominating measure for $(X,Y)$ when $Y=\mathbb I_{X>5}$.

  • $\begingroup$ Unsure what you mean by dominating measure. Why this would imply that the joint density is simply equal to the density $f(x)$? $\endgroup$ Commented Nov 15, 2021 at 21:08
  • $\begingroup$ Because $Y$ conditional on $X$ is a Dirac mass. $\endgroup$
    – Xi'an
    Commented Nov 15, 2021 at 21:25

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