Usage of the equality [tex]P[g(X) \leq y]=P[X \leq h(y)][/tex] in an example

fisher garrry

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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?
 

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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?

As I understand it, you are trying to understand the proof by following through it for a specific example, right?

You didn't say anything about how you got those results of 0.5 and 0.466. Also, why do your integrals start at 0 rather than -infinity? But the main problem is that you made a naive substitution without a proper change of variables (involving the differential as well as the integrand).

The equality you are asking about is "obvious"; it can't not be true! It says that the probabilities of two events are equal, which is true because the events are the same event! Do you see why that is so? Then, "applying" it just means knowing that the two integrals (when you state them correctly and evaluate them correctly) will be equal. What you are doing should confirm that, though it is not necessary.
 
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As I understand it, you are trying to understand the proof by following through it for a specific example, right?

You didn't say anything about how you got those results of 0.5 and 0.466. Also, why do your integrals start at 0 rather than -infinity? But the main problem is that you made a naive substitution without a proper change of variables (involving the differential as well as the integrand).

The equality you are asking about is "obvious"; it can't not be true! It says that the probabilities of two events are equal, which is true because the events are the same event! Do you see why that is so? Then, "applying" it just means knowing that the two integrals (when you state them correctly and evaluate them correctly) will be equal. What you are doing should confirm that, though it is not necessary.

Thanks for the answer. But I still want to work through an example. If anyone can do that it would still help.
 
Thanks for the answer. But I still want to work through an example. If anyone can do that it would still help.
Before that, "work through" your work - correct it and post it. Then we can think about another example and work through (you first and then us).
 
Before that, "work through" your work - correct it and post it. Then we can think about another example and work through (you first and then us).

If I again start with the standard normal distribution for a variable y as the pdf. That is \(\displaystyle \frac{1}{\sqrt{2 \pi}}e^{-\frac{y^2}{2}}\). If we look at the cdf for the pdf for numbers up until a we obtain: \(\displaystyle F_y=P(Y \leq a)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^a e^{-\frac{y^2}{2}} dy=\frac{erf \frac{a}{\sqrt{2}}+1}{2} \)
if \(\displaystyle y=g(x)=x^3\) then \(\displaystyle F_y=\frac{erf \frac{x^3}{\sqrt{2}}+1}{2} \)

But then we also know that \(\displaystyle P(g(x) \leq a]=P(x \leq h(a)) \)

But then what is \(\displaystyle F_x\). We know \(\displaystyle F_y\) but what is \(\displaystyle F_x\)?
 
If I again start with the standard normal distribution for a variable y as the pdf. That is \(\displaystyle \frac{1}{\sqrt{2 \pi}}e^{-\frac{y^2}{2}}\). If we look at the cdf for the pdf for numbers up until a we obtain: \(\displaystyle F_y=P(Y \leq a)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^a e^{-\frac{y^2}{2}} dy=\frac{erf \frac{a}{\sqrt{2}}+1}{2} \)
if \(\displaystyle y=g(x)=x^3\) then \(\displaystyle F_y=\frac{erf \frac{x^3}{\sqrt{2}}+1}{2} \)

But then we also know that \(\displaystyle P(g(x) \leq a]=P(x \leq h(a)) \)

But then what is \(\displaystyle F_x\). We know \(\displaystyle F_y\) but what is \(\displaystyle F_x\)?

As I understand it, you are defining \(\displaystyle F_X\) as \(\displaystyle F_X(x) = P\left(X \leq x\right)\). (You changed from upper case to lower case subscripts in your most recent message, which confused me; I initially thought these were partial derivatives!)

Also, you originally were taking X as standard normal, and now it is Y, which doesn't seem to fit what you were trying to do before.

But what you did initially was close to being correct, apart from the details I mentioned. Try again to make the change of variables in the integral (though my guess is that the result will not be integrable). You have to replace x with its equivalent in terms of y, and likewise replace dx with its equivalent in terms of y and dy. Have you tried doing that?
 
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