fisher garrry
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- Dec 11, 2017
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THEOREM
Let \(\displaystyle X\) have pdf \(\displaystyle f_{X}(x)\) and let \(\displaystyle Y\, =\, g(X),\) where \(\displaystyle g\) is monotonic (either strictly increasing or strictly decreasing) so it has an inverse function \(\displaystyle X\, =\, h(Y).\) Assume that \(\displaystyle h\) has a derivative \(\displaystyle h'(y).\) Then:
. . . . .\(\displaystyle f_Y(y)\, =\, f_X(h(y))\, |h'(y)|\)
Proof (for "strictly increasing" case):
We follow the last method in Example 4.38. First, find the cdf:
. . . . .\(\displaystyle F_Y(y)\, =\, P(Y\, \leq\, y)\, =\, P\left[g(X)\, \leq\, y\right]\, =\, P\left[X\, \leq\, h(y)\right]\, =\, F_X\left[h(y)\right]\)
Differentiate the cdf, letting \(\displaystyle x\, =\, h(y).\)
. . . . .\(\displaystyle f_Y(y)\, =\, \frac{d}{dy}\,F_Y(y)\, =\, \frac{d}{dy}\,F_X\left[h(y)\right]\, =\, \frac{dx\, d}{dy\, dx}\, F_X(x)\, =\, h'(y)\, f_X(x)\, =\, h'(y)\, f_X\left[h(y)\right]\)
My problem:
I want to apply the equality \(\displaystyle P[g(X) \leq y]\, =\, P[X\, \leq\, h(y)]\) in the proof to an example to show its validity. Here is what I have tried:
If we look at the standard normal distribution with \(\displaystyle \mu\,=\,0\) and \(\displaystyle \sigma\,=\,1\). Then we apply integration limits 0 to 10 for \(\displaystyle x\). We have that \(\displaystyle g(X)\,=\,x^3\)
\(\displaystyle \displaystyle \int_{0}^{10}\, \scriptsize{\frac{1}{\sqrt{\strut 2 \pi\,}}}\,\)\(\displaystyle \, e^{-{\frac{x^2}{2}}}\, dx\,=\,0.5\)
and for the inverse \(\displaystyle \sqrt[3]{x\,}\) 0 to \(\displaystyle \sqrt[3]{10\,}\, =\,1.78\)
\(\displaystyle \displaystyle \int_{0}^{1.78} \scriptsize{\frac{1}{\sqrt{2 \pi}}}\)\(\displaystyle \, e^{-{\frac{\sqrt[3]{x}^2}{2}}}\, dx\, =\,0.466\)
But \(\displaystyle 0.5\, \neq\, 0.466\)
What is wrong. Can someone show how I can use the equality \(\displaystyle P[g(X) \leq y]\, =\, P[X\, \leq \, h(y)]\) on this example. And point out what is wrong?
Let \(\displaystyle X\) have pdf \(\displaystyle f_{X}(x)\) and let \(\displaystyle Y\, =\, g(X),\) where \(\displaystyle g\) is monotonic (either strictly increasing or strictly decreasing) so it has an inverse function \(\displaystyle X\, =\, h(Y).\) Assume that \(\displaystyle h\) has a derivative \(\displaystyle h'(y).\) Then:
. . . . .\(\displaystyle f_Y(y)\, =\, f_X(h(y))\, |h'(y)|\)
Proof (for "strictly increasing" case):
We follow the last method in Example 4.38. First, find the cdf:
. . . . .\(\displaystyle F_Y(y)\, =\, P(Y\, \leq\, y)\, =\, P\left[g(X)\, \leq\, y\right]\, =\, P\left[X\, \leq\, h(y)\right]\, =\, F_X\left[h(y)\right]\)
Differentiate the cdf, letting \(\displaystyle x\, =\, h(y).\)
. . . . .\(\displaystyle f_Y(y)\, =\, \frac{d}{dy}\,F_Y(y)\, =\, \frac{d}{dy}\,F_X\left[h(y)\right]\, =\, \frac{dx\, d}{dy\, dx}\, F_X(x)\, =\, h'(y)\, f_X(x)\, =\, h'(y)\, f_X\left[h(y)\right]\)
My problem:
I want to apply the equality \(\displaystyle P[g(X) \leq y]\, =\, P[X\, \leq\, h(y)]\) in the proof to an example to show its validity. Here is what I have tried:
If we look at the standard normal distribution with \(\displaystyle \mu\,=\,0\) and \(\displaystyle \sigma\,=\,1\). Then we apply integration limits 0 to 10 for \(\displaystyle x\). We have that \(\displaystyle g(X)\,=\,x^3\)
\(\displaystyle \displaystyle \int_{0}^{10}\, \scriptsize{\frac{1}{\sqrt{\strut 2 \pi\,}}}\,\)\(\displaystyle \, e^{-{\frac{x^2}{2}}}\, dx\,=\,0.5\)
and for the inverse \(\displaystyle \sqrt[3]{x\,}\) 0 to \(\displaystyle \sqrt[3]{10\,}\, =\,1.78\)
\(\displaystyle \displaystyle \int_{0}^{1.78} \scriptsize{\frac{1}{\sqrt{2 \pi}}}\)\(\displaystyle \, e^{-{\frac{\sqrt[3]{x}^2}{2}}}\, dx\, =\,0.466\)
But \(\displaystyle 0.5\, \neq\, 0.466\)
What is wrong. Can someone show how I can use the equality \(\displaystyle P[g(X) \leq y]\, =\, P[X\, \leq \, h(y)]\) on this example. And point out what is wrong?
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