Let $X$ be a random variable with CDF $F_X(x)$ given by $$ F_X(x)=1-\frac{\Gamma(m,(m/y)x)}{\Gamma(m)}, $$

where $m$ and $y$ are positive integers $(m>0, y>0)$ and $\Gamma(a,z)$ is the incomplete gamma function defined $$ \Gamma(a,z)=\int_{z}^{\infty}t^{a-1}e^{-t}dt. $$ How we can find the PDF of $X$, $f_X(x)$ using derivation method?.

The PDF of $X$ is given by $$ f_X(x)=\frac{m^m}{y^m\Gamma(m)}x^{m-1}e^{-(m/y)x}. $$

My quetion how to get $f_X(x)$ using $$ f_X(x)=\frac{dF_X(x)}{dx}. $$

  • $\begingroup$ What have you tried? Have you, for instance, tried the definition of the derivative (perhaps with Leibniz integral rule)? If so, how much progress did you make? $\endgroup$ – Eric Towers Jan 5 '19 at 22:27
  • $\begingroup$ I would like to find the PDF of $X$ using $f_X(x)=\frac{dF_X(x)}{dx}$. Thanks $\endgroup$ – Monir Jan 5 '19 at 22:31
  • $\begingroup$ You are more likely to get responses if you address the instructions to the right of the box you typed your question into, including "Provide details. Share your research." Which includes "Include your work". $\endgroup$ – Eric Towers Jan 5 '19 at 22:32
  • $\begingroup$ Actually, you seem to be merely after the basic differentiation formula $$\frac d{dx}\int_{cx}^\infty g(t)dt=-cg(cx)$$ $\endgroup$ – Did Jan 6 '19 at 13:26
  • $\begingroup$ OK, this low give the right answer, thanks, could you please provide for me reference. Thanks $\endgroup$ – Monir Jan 6 '19 at 13:28

You have $$F_X(x)=1-\frac{\Gamma(m,(m/y)x)}{\Gamma(m)},$$ where $$\Gamma(a,z) = \int_{z}^{\infty}t^{a-1}e^{-z}dt.$$ Then $$f_X(x)=\frac{dF_{X}(x)}{dx} = \frac{-1}{\Gamma(m)}\left(\int_{(m/y)x}^{\infty}t^{m-1}e^{-(m/y)x}dt\right)'.$$ Now $$\frac{d\Gamma(a,z)}{dz} = -e^{-z}z^{a-1}$$ and so $$f_X(x) = \frac{-1}{\Gamma(m)}\left(\int_{(m/y)x}^{\infty}t^{m-1}e^{-(m/y)x}dt\right)'$$ $$ = \frac{1}{\Gamma(m)}\left[ e^{-(m/y)x}((m/y)x)^{m-1}\right]\frac{m}{y} = \frac{m^m}{y^m\Gamma(m)}x^{m-1}e^{-(m/y)x}.$$

  • $\begingroup$ Please, how did you simplifay $$\frac{-1}{\Gamma(m)}\left(\int_{(m/y)x}^{\infty}t^{m-1}e^{-(m/y)x}dt\right)'$$ $\endgroup$ – Monir Jan 6 '19 at 0:02
  • $\begingroup$ Use en.wikipedia.org/wiki/… $\endgroup$ – nls Jan 6 '19 at 0:03
  • $\begingroup$ D you mean we first derivation then integral? $\endgroup$ – Monir Jan 6 '19 at 0:07
  • $\begingroup$ You want to derivate a function which is written as an integral. You have something like $\Gamma(a,z) = \int_{\alpha}^{\beta} G(t,z)dt.$ $\endgroup$ – nls Jan 6 '19 at 0:09
  • $\begingroup$ ok, just this step $$\frac{-1}{\Gamma(m)}\left(\int_{(m/y)x}^{\infty}t^{m-1}e^{-(m/y)x}dt\right)'$$, could you please add more detail. $\endgroup$ – Monir Jan 6 '19 at 0:11

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