Support for transformation of Standard Normal distribution PDF
I was studying probability distribution functions and had a particular question about an exercise problem. The problem goes:
Let $Z$ ~ $mathcal{N}(0, 1)$. Find the PDF for $Z^4$.
I've solved the problem, but missed defining the support, and frankly am not sure how the support came to be. Allow me to elaborate.
My Solution
Let $Y = Z^4$. We cannot use the change of variables formula due to the fact that $y = z^4$ is not strictly increasing or decreasing. We will find the CDF for $Y$ first, then differentiate to find the PDF.
begin{align}
F_Y(y) & = P(Yle y) \
& = P(Z^4 le y) \
& = P(|Z| le y^{frac{1}{4}}) \
& = P(-y^{frac{1}{4}} le Z le y^{frac{1}{4}}) \
& = P(Z le y^{frac{1}{4}}) - P(Zle -y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - Phi(-y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - (1 - Phi(y^{frac{1}{4}})) \
& = 2Phi(y^{frac{1}{4}}) - 1
end{align}
Now that we've found the CDF of $Y$, we can find the PDF of $Y$ by differentiating the CDF and using the chain rule.
begin{align}
frac{d}{dy}F_Y(y) & = frac{d}{dy}(2Phi(y^{frac{1}{4}}) - 1) \
& = 2frac{d}{dy}Phi(y^{frac{1}{4}}) - frac{d}{dy}1 \
& = 2phi(y^{frac{1}{4}})times frac{d}{dy}(y^{frac{1}{4}}) - 0\
& = frac{1}{4sqrt{2pi}}e^{-frac{y^{1/2}}{2}}y^{-frac{3}{4}}
end{align}
Apparently the equation itself is correct, but I got it wrong because I forgot to specify that it only holds when $y gt 0$ and is $0$ elsewhere.
I'm having trouble how this support came to be. I know that a Normal distribution's support is defined on $(-infty, infty)$, and I guess I was under the assumption that the same would hold for $mathcal{N}^4$. Where did the support for $y$, $(0, infty)$, come from?
Any feedback is appreciated. Thank you.
probability probability-distributions normal-distribution
asked Nov 26 at 14:55
Seankala
24210
add a comment |
I was studying probability distribution functions and had a particular question about an exercise problem. The problem goes:
Let $Z$ ~ $mathcal{N}(0, 1)$. Find the PDF for $Z^4$.
I've solved the problem, but missed defining the support, and frankly am not sure how the support came to be. Allow me to elaborate.
My Solution
Let $Y = Z^4$. We cannot use the change of variables formula due to the fact that $y = z^4$ is not strictly increasing or decreasing. We will find the CDF for $Y$ first, then differentiate to find the PDF.
begin{align}
F_Y(y) & = P(Yle y) \
& = P(Z^4 le y) \
& = P(|Z| le y^{frac{1}{4}}) \
& = P(-y^{frac{1}{4}} le Z le y^{frac{1}{4}}) \
& = P(Z le y^{frac{1}{4}}) - P(Zle -y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - Phi(-y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - (1 - Phi(y^{frac{1}{4}})) \
& = 2Phi(y^{frac{1}{4}}) - 1
end{align}
Now that we've found the CDF of $Y$, we can find the PDF of $Y$ by differentiating the CDF and using the chain rule.
begin{align}
frac{d}{dy}F_Y(y) & = frac{d}{dy}(2Phi(y^{frac{1}{4}}) - 1) \
& = 2frac{d}{dy}Phi(y^{frac{1}{4}}) - frac{d}{dy}1 \
& = 2phi(y^{frac{1}{4}})times frac{d}{dy}(y^{frac{1}{4}}) - 0\
& = frac{1}{4sqrt{2pi}}e^{-frac{y^{1/2}}{2}}y^{-frac{3}{4}}
end{align}
Apparently the equation itself is correct, but I got it wrong because I forgot to specify that it only holds when $y gt 0$ and is $0$ elsewhere.
I'm having trouble how this support came to be. I know that a Normal distribution's support is defined on $(-infty, infty)$, and I guess I was under the assumption that the same would hold for $mathcal{N}^4$. Where did the support for $y$, $(0, infty)$, come from?
Any feedback is appreciated. Thank you.
probability probability-distributions normal-distribution
asked Nov 26 at 14:55
Seankala
24210
1
$zinmathbb Rimplies z^2ge 0implies (z^2)^2ge 0$.
– StubbornAtom
Nov 26 at 14:58
Thank you! You may also post an answer if you'd like. :)
– Seankala
Nov 26 at 15:40
add a comment |
I was studying probability distribution functions and had a particular question about an exercise problem. The problem goes:
Let $Z$ ~ $mathcal{N}(0, 1)$. Find the PDF for $Z^4$.
I've solved the problem, but missed defining the support, and frankly am not sure how the support came to be. Allow me to elaborate.
My Solution
Let $Y = Z^4$. We cannot use the change of variables formula due to the fact that $y = z^4$ is not strictly increasing or decreasing. We will find the CDF for $Y$ first, then differentiate to find the PDF.
begin{align}
F_Y(y) & = P(Yle y) \
& = P(Z^4 le y) \
& = P(|Z| le y^{frac{1}{4}}) \
& = P(-y^{frac{1}{4}} le Z le y^{frac{1}{4}}) \
& = P(Z le y^{frac{1}{4}}) - P(Zle -y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - Phi(-y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - (1 - Phi(y^{frac{1}{4}})) \
& = 2Phi(y^{frac{1}{4}}) - 1
end{align}
Now that we've found the CDF of $Y$, we can find the PDF of $Y$ by differentiating the CDF and using the chain rule.
begin{align}
frac{d}{dy}F_Y(y) & = frac{d}{dy}(2Phi(y^{frac{1}{4}}) - 1) \
& = 2frac{d}{dy}Phi(y^{frac{1}{4}}) - frac{d}{dy}1 \
& = 2phi(y^{frac{1}{4}})times frac{d}{dy}(y^{frac{1}{4}}) - 0\
& = frac{1}{4sqrt{2pi}}e^{-frac{y^{1/2}}{2}}y^{-frac{3}{4}}
end{align}
Apparently the equation itself is correct, but I got it wrong because I forgot to specify that it only holds when $y gt 0$ and is $0$ elsewhere.
I'm having trouble how this support came to be. I know that a Normal distribution's support is defined on $(-infty, infty)$, and I guess I was under the assumption that the same would hold for $mathcal{N}^4$. Where did the support for $y$, $(0, infty)$, come from?
Any feedback is appreciated. Thank you.
probability probability-distributions normal-distribution
asked Nov 26 at 14:55
Seankala
24210
I was studying probability distribution functions and had a particular question about an exercise problem. The problem goes:
Let $Z$ ~ $mathcal{N}(0, 1)$. Find the PDF for $Z^4$.
I've solved the problem, but missed defining the support, and frankly am not sure how the support came to be. Allow me to elaborate.
My Solution
Let $Y = Z^4$. We cannot use the change of variables formula due to the fact that $y = z^4$ is not strictly increasing or decreasing. We will find the CDF for $Y$ first, then differentiate to find the PDF.
begin{align}
F_Y(y) & = P(Yle y) \
& = P(Z^4 le y) \
& = P(|Z| le y^{frac{1}{4}}) \
& = P(-y^{frac{1}{4}} le Z le y^{frac{1}{4}}) \
& = P(Z le y^{frac{1}{4}}) - P(Zle -y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - Phi(-y^{frac{1}{4}}) \
& = Phi(y^{frac{1}{4}}) - (1 - Phi(y^{frac{1}{4}})) \
& = 2Phi(y^{frac{1}{4}}) - 1
end{align}
Now that we've found the CDF of $Y$, we can find the PDF of $Y$ by differentiating the CDF and using the chain rule.
begin{align}
frac{d}{dy}F_Y(y) & = frac{d}{dy}(2Phi(y^{frac{1}{4}}) - 1) \
& = 2frac{d}{dy}Phi(y^{frac{1}{4}}) - frac{d}{dy}1 \
& = 2phi(y^{frac{1}{4}})times frac{d}{dy}(y^{frac{1}{4}}) - 0\
& = frac{1}{4sqrt{2pi}}e^{-frac{y^{1/2}}{2}}y^{-frac{3}{4}}
end{align}
Apparently the equation itself is correct, but I got it wrong because I forgot to specify that it only holds when $y gt 0$ and is $0$ elsewhere.
I'm having trouble how this support came to be. I know that a Normal distribution's support is defined on $(-infty, infty)$, and I guess I was under the assumption that the same would hold for $mathcal{N}^4$. Where did the support for $y$, $(0, infty)$, come from?
Any feedback is appreciated. Thank you.
probability probability-distributions normal-distribution
probability probability-distributions normal-distribution
asked Nov 26 at 14:55
Seankala
24210
asked Nov 26 at 14:55
Seankala
24210
asked Nov 26 at 14:55
Seankala
24210
asked Nov 26 at 14:55
Seankala
24210
asked Nov 26 at 14:55
Seankala
24210
24210
1
$zinmathbb Rimplies z^2ge 0implies (z^2)^2ge 0$.
– StubbornAtom
Nov 26 at 14:58
Thank you! You may also post an answer if you'd like. :)
– Seankala
Nov 26 at 15:40
add a comment |
1
$zinmathbb Rimplies z^2ge 0implies (z^2)^2ge 0$.
– StubbornAtom
Nov 26 at 14:58
Thank you! You may also post an answer if you'd like. :)
– Seankala
Nov 26 at 15:40
1
1
$zinmathbb Rimplies z^2ge 0implies (z^2)^2ge 0$.
– StubbornAtom
Nov 26 at 14:58
$zinmathbb Rimplies z^2ge 0implies (z^2)^2ge 0$.
– StubbornAtom
Nov 26 at 14:58
Thank you! You may also post an answer if you'd like. :)
– Seankala
Nov 26 at 15:40
Thank you! You may also post an answer if you'd like. :)
– Seankala
Nov 26 at 15:40
add a comment |
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1
$zinmathbb Rimplies z^2ge 0implies (z^2)^2ge 0$.
– StubbornAtom
Nov 26 at 14:58
Thank you! You may also post an answer if you'd like. :)
– Seankala
Nov 26 at 15:40