Find the PDF of gamma distributed random variable using derivation
$begingroup$
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
asked Jan 5 at 22:12
MonirMonir
328
$endgroup$
add a comment |
$begingroup$
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
asked Jan 5 at 22:12
MonirMonir
328
$endgroup$
$begingroup$
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 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 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 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 at 13:26
$begingroup$
OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
$endgroup$
– Monir
Jan 6 at 13:28
add a comment |
$begingroup$
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
asked Jan 5 at 22:12
MonirMonir
328
$endgroup$
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
asked Jan 5 at 22:12
MonirMonir
328
asked Jan 5 at 22:12
MonirMonir
328
edited Jan 6 at 13:20
Monir
asked Jan 5 at 22:12
MonirMonir
328
asked Jan 5 at 22:12
MonirMonir
328
asked Jan 5 at 22:12
MonirMonir
328
328
$begingroup$
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 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 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 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 at 13:26
$begingroup$
OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
$endgroup$
– Monir
Jan 6 at 13:28
add a comment |
$begingroup$
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 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 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 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 at 13:26
$begingroup$
OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
$endgroup$
– Monir
Jan 6 at 13:28
$begingroup$
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 at 22:27
$begingroup$
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 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 at 22:31
$begingroup$
I would like to find the PDF of $X$ using $f_X(x)=frac{dF_X(x)}{dx}$. Thanks
$endgroup$
– Monir
Jan 5 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 at 22:32
$begingroup$
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 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 at 13:26
$begingroup$
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 at 13:26
$begingroup$
OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
$endgroup$
– Monir
Jan 6 at 13:28
$begingroup$
OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
$endgroup$
– Monir
Jan 6 at 13:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
$endgroup$
$begingroup$
Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$endgroup$
– Monir
Jan 6 at 0:02
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– Hello_World
Jan 6 at 0:03
$begingroup$
D you mean we first derivation then integral?
$endgroup$
– Monir
Jan 6 at 0:07
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– Hello_World
Jan 6 at 0:09
$begingroup$
ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
$endgroup$
– Monir
Jan 6 at 0:11
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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$begingroup$
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
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
$endgroup$
$begingroup$
Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$endgroup$
– Monir
Jan 6 at 0:02
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– Hello_World
Jan 6 at 0:03
$begingroup$
D you mean we first derivation then integral?
$endgroup$
– Monir
Jan 6 at 0:07
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– Hello_World
Jan 6 at 0:09
$begingroup$
ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
$endgroup$
– Monir
Jan 6 at 0:11
|
show 2 more comments
$begingroup$
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
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
$endgroup$
$begingroup$
Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$endgroup$
– Monir
Jan 6 at 0:02
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– Hello_World
Jan 6 at 0:03
$begingroup$
D you mean we first derivation then integral?
$endgroup$
– Monir
Jan 6 at 0:07
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– Hello_World
Jan 6 at 0:09
$begingroup$
ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
$endgroup$
– Monir
Jan 6 at 0:11
|
show 2 more comments
$begingroup$
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
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
$endgroup$
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
$$f_X(x) = frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$$ = frac{1}{Gamma(m)}left[ e^{-(m/y)x}((m/y)x)^{m-1}right]frac{m}{y} = frac{m^m}{y^mGamma(m)}x^{m-1}e^{-(m/y)x}.$$
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
answered Jan 5 at 23:40
Hello_WorldHello_World
4,14621931
4,14621931
$begingroup$
Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$endgroup$
– Monir
Jan 6 at 0:02
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– Hello_World
Jan 6 at 0:03
$begingroup$
D you mean we first derivation then integral?
$endgroup$
– Monir
Jan 6 at 0:07
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– Hello_World
Jan 6 at 0:09
$begingroup$
ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
$endgroup$
– Monir
Jan 6 at 0:11
|
show 2 more comments
$begingroup$
Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$endgroup$
– Monir
Jan 6 at 0:02
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– Hello_World
Jan 6 at 0:03
$begingroup$
D you mean we first derivation then integral?
$endgroup$
– Monir
Jan 6 at 0:07
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– Hello_World
Jan 6 at 0:09
$begingroup$
ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
$endgroup$
– Monir
Jan 6 at 0:11
$begingroup$
Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$endgroup$
– Monir
Jan 6 at 0:02
$begingroup$
Please, how did you simplifay $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$
$endgroup$
– Monir
Jan 6 at 0:02
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– Hello_World
Jan 6 at 0:03
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– Hello_World
Jan 6 at 0:03
$begingroup$
D you mean we first derivation then integral?
$endgroup$
– Monir
Jan 6 at 0:07
$begingroup$
D you mean we first derivation then integral?
$endgroup$
– Monir
Jan 6 at 0:07
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– Hello_World
Jan 6 at 0:09
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– Hello_World
Jan 6 at 0:09
$begingroup$
ok, just this step $$frac{-1}{Gamma(m)}left(int_{(m/y)x}^{infty}t^{m-1}e^{-(m/y)x}dtright)'$$, could you please add more detail.
$endgroup$
– Monir
Jan 6 at 0:11
$begingroup$
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.
$endgroup$
– Monir
Jan 6 at 0:11
|
show 2 more comments
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$begingroup$
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 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 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 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 at 13:26
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OK, this low give the right answer, thanks, could you please provide for me reference. Thanks
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– Monir
Jan 6 at 13:28