Derive the PDF of the log-normal distribution?












1












$begingroup$


If $X sim N(0,1)$ and $Y = e^X$, find the PDF of $Y$ using the two methods:


(i) Find the CDF of of $Y$ and then differentiate. Use the notation $Phi(x)$ and $phi(x)$ for the CDF and PDF of $X$ respectively. You may use the fact that $phi(x) = Phi'(x)$.


So I'm not sure how to differentiate $Phibig(dfrac{ln x-mu}{sigma} bigg)$ to get $dfrac{1}{xsigmasqrt{2pi}}e^-frac{(ln x-mu)^2}{2sigma^2}$


(ii) Use the transformation formula.


I'm not sure where to even begin with this one.






statistics probability-distributions logarithms normal-distribution exponential-function







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$endgroup$












  • $begingroup$
    For (i), just use the chain rule for differentiation, remembering that $$frac{d}{dx}Phi(g(x))=phi(g(x))g'(x) ~~mathrm{where}~~ phi(y)=frac{1}{sqrt{2pi}}e^{-frac{y^2}{2}}.$$
    $endgroup$
    – Dilip Sarwate
    Apr 23 '13 at 3:12
















1












$begingroup$


If $X sim N(0,1)$ and $Y = e^X$, find the PDF of $Y$ using the two methods:


(i) Find the CDF of of $Y$ and then differentiate. Use the notation $Phi(x)$ and $phi(x)$ for the CDF and PDF of $X$ respectively. You may use the fact that $phi(x) = Phi'(x)$.


So I'm not sure how to differentiate $Phibig(dfrac{ln x-mu}{sigma} bigg)$ to get $dfrac{1}{xsigmasqrt{2pi}}e^-frac{(ln x-mu)^2}{2sigma^2}$


(ii) Use the transformation formula.


I'm not sure where to even begin with this one.






statistics probability-distributions logarithms normal-distribution exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    For (i), just use the chain rule for differentiation, remembering that $$frac{d}{dx}Phi(g(x))=phi(g(x))g'(x) ~~mathrm{where}~~ phi(y)=frac{1}{sqrt{2pi}}e^{-frac{y^2}{2}}.$$
    $endgroup$
    – Dilip Sarwate
    Apr 23 '13 at 3:12














1












1








1


2



$begingroup$


If $X sim N(0,1)$ and $Y = e^X$, find the PDF of $Y$ using the two methods:


(i) Find the CDF of of $Y$ and then differentiate. Use the notation $Phi(x)$ and $phi(x)$ for the CDF and PDF of $X$ respectively. You may use the fact that $phi(x) = Phi'(x)$.


So I'm not sure how to differentiate $Phibig(dfrac{ln x-mu}{sigma} bigg)$ to get $dfrac{1}{xsigmasqrt{2pi}}e^-frac{(ln x-mu)^2}{2sigma^2}$


(ii) Use the transformation formula.


I'm not sure where to even begin with this one.






statistics probability-distributions logarithms normal-distribution exponential-function







share|cite|improve this question











$endgroup$




If $X sim N(0,1)$ and $Y = e^X$, find the PDF of $Y$ using the two methods:


(i) Find the CDF of of $Y$ and then differentiate. Use the notation $Phi(x)$ and $phi(x)$ for the CDF and PDF of $X$ respectively. You may use the fact that $phi(x) = Phi'(x)$.


So I'm not sure how to differentiate $Phibig(dfrac{ln x-mu}{sigma} bigg)$ to get $dfrac{1}{xsigmasqrt{2pi}}e^-frac{(ln x-mu)^2}{2sigma^2}$


(ii) Use the transformation formula.


I'm not sure where to even begin with this one.






statistics probability-distributions logarithms normal-distribution exponential-function




statistics probability-distributions logarithms normal-distribution exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 22 '13 at 20:20









Alex

14.3k42134




14.3k42134










asked Apr 22 '13 at 20:14









MathleteMathlete

6273923




6273923












  • $begingroup$
    For (i), just use the chain rule for differentiation, remembering that $$frac{d}{dx}Phi(g(x))=phi(g(x))g'(x) ~~mathrm{where}~~ phi(y)=frac{1}{sqrt{2pi}}e^{-frac{y^2}{2}}.$$
    $endgroup$
    – Dilip Sarwate
    Apr 23 '13 at 3:12


















  • $begingroup$
    For (i), just use the chain rule for differentiation, remembering that $$frac{d}{dx}Phi(g(x))=phi(g(x))g'(x) ~~mathrm{where}~~ phi(y)=frac{1}{sqrt{2pi}}e^{-frac{y^2}{2}}.$$
    $endgroup$
    – Dilip Sarwate
    Apr 23 '13 at 3:12
















$begingroup$
For (i), just use the chain rule for differentiation, remembering that $$frac{d}{dx}Phi(g(x))=phi(g(x))g'(x) ~~mathrm{where}~~ phi(y)=frac{1}{sqrt{2pi}}e^{-frac{y^2}{2}}.$$
$endgroup$
– Dilip Sarwate
Apr 23 '13 at 3:12




$begingroup$
For (i), just use the chain rule for differentiation, remembering that $$frac{d}{dx}Phi(g(x))=phi(g(x))g'(x) ~~mathrm{where}~~ phi(y)=frac{1}{sqrt{2pi}}e^{-frac{y^2}{2}}.$$
$endgroup$
– Dilip Sarwate
Apr 23 '13 at 3:12










1 Answer
1






active

oldest

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0












$begingroup$

If $Y=e^X$, then $varphi^{-1}(Y)= log Y$. Hence, $f_X (y) =frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}$ and differentiate $|frac{d varphi^{-1}(Y)}{dy} | = |frac{1}{Y}|$. Hence, the pdf of $Y$ is
$$
h_{Y}(y) = frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}frac{1}{|y|}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is called function of the single random variable, that can probably be referred to as transformation.
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression?
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:48










  • $begingroup$
    Just log both sides. Then take a derivative wrt to $y$
    $endgroup$
    – Alex
    Apr 22 '13 at 20:49










  • $begingroup$
    I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with.
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:51











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1 Answer
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1 Answer
1






active

oldest

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0












$begingroup$

If $Y=e^X$, then $varphi^{-1}(Y)= log Y$. Hence, $f_X (y) =frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}$ and differentiate $|frac{d varphi^{-1}(Y)}{dy} | = |frac{1}{Y}|$. Hence, the pdf of $Y$ is
$$
h_{Y}(y) = frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}frac{1}{|y|}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is called function of the single random variable, that can probably be referred to as transformation.
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression?
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:48










  • $begingroup$
    Just log both sides. Then take a derivative wrt to $y$
    $endgroup$
    – Alex
    Apr 22 '13 at 20:49










  • $begingroup$
    I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with.
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:51
















0












$begingroup$

If $Y=e^X$, then $varphi^{-1}(Y)= log Y$. Hence, $f_X (y) =frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}$ and differentiate $|frac{d varphi^{-1}(Y)}{dy} | = |frac{1}{Y}|$. Hence, the pdf of $Y$ is
$$
h_{Y}(y) = frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}frac{1}{|y|}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is called function of the single random variable, that can probably be referred to as transformation.
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression?
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:48










  • $begingroup$
    Just log both sides. Then take a derivative wrt to $y$
    $endgroup$
    – Alex
    Apr 22 '13 at 20:49










  • $begingroup$
    I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with.
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:51














0












0








0





$begingroup$

If $Y=e^X$, then $varphi^{-1}(Y)= log Y$. Hence, $f_X (y) =frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}$ and differentiate $|frac{d varphi^{-1}(Y)}{dy} | = |frac{1}{Y}|$. Hence, the pdf of $Y$ is
$$
h_{Y}(y) = frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}frac{1}{|y|}
$$






share|cite|improve this answer









$endgroup$



If $Y=e^X$, then $varphi^{-1}(Y)= log Y$. Hence, $f_X (y) =frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}$ and differentiate $|frac{d varphi^{-1}(Y)}{dy} | = |frac{1}{Y}|$. Hence, the pdf of $Y$ is
$$
h_{Y}(y) = frac{1}{sqrt{2 pi}}e^{-frac{log^2 y}{2}}frac{1}{|y|}
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 22 '13 at 20:36









AlexAlex

14.3k42134




14.3k42134












  • $begingroup$
    This is called function of the single random variable, that can probably be referred to as transformation.
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression?
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:48










  • $begingroup$
    Just log both sides. Then take a derivative wrt to $y$
    $endgroup$
    – Alex
    Apr 22 '13 at 20:49










  • $begingroup$
    I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with.
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:51


















  • $begingroup$
    This is called function of the single random variable, that can probably be referred to as transformation.
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Alex
    Apr 22 '13 at 20:48










  • $begingroup$
    Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression?
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:48










  • $begingroup$
    Just log both sides. Then take a derivative wrt to $y$
    $endgroup$
    – Alex
    Apr 22 '13 at 20:49










  • $begingroup$
    I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with.
    $endgroup$
    – Mathlete
    Apr 22 '13 at 20:51
















$begingroup$
This is called function of the single random variable, that can probably be referred to as transformation.
$endgroup$
– Alex
Apr 22 '13 at 20:48




$begingroup$
This is called function of the single random variable, that can probably be referred to as transformation.
$endgroup$
– Alex
Apr 22 '13 at 20:48












$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Alex
Apr 22 '13 at 20:48




$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Alex
Apr 22 '13 at 20:48












$begingroup$
Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression?
$endgroup$
– Mathlete
Apr 22 '13 at 20:48




$begingroup$
Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression?
$endgroup$
– Mathlete
Apr 22 '13 at 20:48












$begingroup$
Just log both sides. Then take a derivative wrt to $y$
$endgroup$
– Alex
Apr 22 '13 at 20:49




$begingroup$
Just log both sides. Then take a derivative wrt to $y$
$endgroup$
– Alex
Apr 22 '13 at 20:49












$begingroup$
I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with.
$endgroup$
– Mathlete
Apr 22 '13 at 20:51




$begingroup$
I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with.
$endgroup$
– Mathlete
Apr 22 '13 at 20:51


















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