Find the PDF of gamma distributed random variable using derivation

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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
$$
f_X(x)=frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.
$$



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






gamma-function gamma-distribution







share|cite|improve this question
























  • 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?
    – Eric Towers
    Jan 5 at 22:27










  • I would like to find the PDF of $X$ using $f_X(x)=frac{dF_X(x)}{dx}$. Thanks
    – Monir
    Jan 5 at 22:31












  • 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".
    – Eric Towers
    Jan 5 at 22:32










  • Actually, you seem to be merely after the basic differentiation formula $$frac d{dx}int_{cx}^infty g(t)dt=-cg(cx)$$
    – Did
    2 days ago












  • OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
    – Monir
    2 days ago
















0














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
$$
f_X(x)=frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.
$$



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






gamma-function gamma-distribution







share|cite|improve this question
























  • 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?
    – Eric Towers
    Jan 5 at 22:27










  • I would like to find the PDF of $X$ using $f_X(x)=frac{dF_X(x)}{dx}$. Thanks
    – Monir
    Jan 5 at 22:31












  • 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".
    – Eric Towers
    Jan 5 at 22:32










  • Actually, you seem to be merely after the basic differentiation formula $$frac d{dx}int_{cx}^infty g(t)dt=-cg(cx)$$
    – Did
    2 days ago












  • OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
    – Monir
    2 days ago














0












0








0







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
$$
f_X(x)=frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.
$$



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






gamma-function gamma-distribution







share|cite|improve this question















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
$$
f_X(x)=frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.
$$



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






gamma-function gamma-distribution




gamma-function gamma-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







Monir

















asked Jan 5 at 22:12









MonirMonir

157




157












  • 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?
    – Eric Towers
    Jan 5 at 22:27










  • I would like to find the PDF of $X$ using $f_X(x)=frac{dF_X(x)}{dx}$. Thanks
    – Monir
    Jan 5 at 22:31












  • 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".
    – Eric Towers
    Jan 5 at 22:32










  • Actually, you seem to be merely after the basic differentiation formula $$frac d{dx}int_{cx}^infty g(t)dt=-cg(cx)$$
    – Did
    2 days ago












  • OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
    – Monir
    2 days ago


















  • 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?
    – Eric Towers
    Jan 5 at 22:27










  • I would like to find the PDF of $X$ using $f_X(x)=frac{dF_X(x)}{dx}$. Thanks
    – Monir
    Jan 5 at 22:31












  • 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".
    – Eric Towers
    Jan 5 at 22:32










  • Actually, you seem to be merely after the basic differentiation formula $$frac d{dx}int_{cx}^infty g(t)dt=-cg(cx)$$
    – Did
    2 days ago












  • OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
    – Monir
    2 days ago
















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?
– Eric Towers
Jan 5 at 22:27




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?
– Eric Towers
Jan 5 at 22:27












I would like to find the PDF of $X$ using $f_X(x)=frac{dF_X(x)}{dx}$. Thanks
– Monir
Jan 5 at 22:31






I would like to find the PDF of $X$ using $f_X(x)=frac{dF_X(x)}{dx}$. Thanks
– Monir
Jan 5 at 22:31














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".
– Eric Towers
Jan 5 at 22:32




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".
– Eric Towers
Jan 5 at 22:32












Actually, you seem to be merely after the basic differentiation formula $$frac d{dx}int_{cx}^infty g(t)dt=-cg(cx)$$
– Did
2 days ago






Actually, you seem to be merely after the basic differentiation formula $$frac d{dx}int_{cx}^infty g(t)dt=-cg(cx)$$
– Did
2 days ago














OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
– Monir
2 days ago




OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
– Monir
2 days ago










1 Answer
1






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oldest

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0














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
$$f_X(x)=frac{dF_{X}(x)}{dx} = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'.$$
Now
$$frac{dGamma(a,z)}{dz} = -e^{-z}z^{a-1}$$
and so
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$






share|cite|improve this answer





















  • Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
    – Monir
    Jan 6 at 0:02










  • Use en.wikipedia.org/wiki/…
    – Hello_World
    Jan 6 at 0:03










  • D you mean we first derivation then integral?
    – Monir
    Jan 6 at 0:07










  • You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
    – Hello_World
    Jan 6 at 0:09










  • ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
    – Monir
    Jan 6 at 0:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














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
$$f_X(x)=frac{dF_{X}(x)}{dx} = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'.$$
Now
$$frac{dGamma(a,z)}{dz} = -e^{-z}z^{a-1}$$
and so
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$






share|cite|improve this answer





















  • Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
    – Monir
    Jan 6 at 0:02










  • Use en.wikipedia.org/wiki/…
    – Hello_World
    Jan 6 at 0:03










  • D you mean we first derivation then integral?
    – Monir
    Jan 6 at 0:07










  • You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
    – Hello_World
    Jan 6 at 0:09










  • ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
    – Monir
    Jan 6 at 0:11
















0














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
$$f_X(x)=frac{dF_{X}(x)}{dx} = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'.$$
Now
$$frac{dGamma(a,z)}{dz} = -e^{-z}z^{a-1}$$
and so
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$






share|cite|improve this answer





















  • Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
    – Monir
    Jan 6 at 0:02










  • Use en.wikipedia.org/wiki/…
    – Hello_World
    Jan 6 at 0:03










  • D you mean we first derivation then integral?
    – Monir
    Jan 6 at 0:07










  • You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
    – Hello_World
    Jan 6 at 0:09










  • ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
    – Monir
    Jan 6 at 0:11














0












0








0






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
$$f_X(x)=frac{dF_{X}(x)}{dx} = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'.$$
Now
$$frac{dGamma(a,z)}{dz} = -e^{-z}z^{a-1}$$
and so
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$






share|cite|improve this answer












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
$$f_X(x)=frac{dF_{X}(x)}{dx} = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'.$$
Now
$$frac{dGamma(a,z)}{dz} = -e^{-z}z^{a-1}$$
and so
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 23:40









Hello_WorldHello_World

4,17321630




4,17321630












  • Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
    – Monir
    Jan 6 at 0:02










  • Use en.wikipedia.org/wiki/…
    – Hello_World
    Jan 6 at 0:03










  • D you mean we first derivation then integral?
    – Monir
    Jan 6 at 0:07










  • You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
    – Hello_World
    Jan 6 at 0:09










  • ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
    – Monir
    Jan 6 at 0:11


















  • Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
    – Monir
    Jan 6 at 0:02










  • Use en.wikipedia.org/wiki/…
    – Hello_World
    Jan 6 at 0:03










  • D you mean we first derivation then integral?
    – Monir
    Jan 6 at 0:07










  • You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
    – Hello_World
    Jan 6 at 0:09










  • ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
    – Monir
    Jan 6 at 0:11
















Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
– Monir
Jan 6 at 0:02




Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
– Monir
Jan 6 at 0:02












Use en.wikipedia.org/wiki/…
– Hello_World
Jan 6 at 0:03




Use en.wikipedia.org/wiki/…
– Hello_World
Jan 6 at 0:03












D you mean we first derivation then integral?
– Monir
Jan 6 at 0:07




D you mean we first derivation then integral?
– Monir
Jan 6 at 0:07












You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
– Hello_World
Jan 6 at 0:09




You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
– Hello_World
Jan 6 at 0:09












ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
– Monir
Jan 6 at 0:11




ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
– Monir
Jan 6 at 0:11


















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