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
539
$endgroup$
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
539
$endgroup$
$begingroup$
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
539
$endgroup$
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
539
asked Jan 5 at 22:12
MonirMonir
539
edited Jan 6 at 13:20
Monir
asked Jan 5 at 22:12
MonirMonir
539
asked Jan 5 at 22:12
MonirMonir
539
asked Jan 5 at 22:12
MonirMonir
539
539
$begingroup$
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
model_checkermodel_checker
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/…
$endgroup$
– model_checker
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 $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– model_checker
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
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063250%2ffind-the-pdf-of-gamma-distributed-random-variable-using-derivation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
model_checkermodel_checker
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/…
$endgroup$
– model_checker
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 $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– model_checker
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
model_checkermodel_checker
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/…
$endgroup$
– model_checker
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 $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– model_checker
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
model_checkermodel_checker
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
model_checkermodel_checker
4,14621931
answered Jan 5 at 23:40
model_checkermodel_checker
4,14621931
answered Jan 5 at 23:40
model_checkermodel_checker
4,14621931
answered Jan 5 at 23:40
model_checkermodel_checker
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/…
$endgroup$
– model_checker
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 $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– model_checker
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/…
$endgroup$
– model_checker
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 $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– model_checker
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/…
$endgroup$
– model_checker
Jan 6 at 0:03
$begingroup$
Use en.wikipedia.org/wiki/…
$endgroup$
– model_checker
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 $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– model_checker
Jan 6 at 0:09
$begingroup$
You want to derivate a function which is written as an integral. You have something like $Gamma(a,z) = int_{alpha}^{beta} G(t,z)dt.$
$endgroup$
– model_checker
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
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063250%2ffind-the-pdf-of-gamma-distributed-random-variable-using-derivation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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