Uniform Distribution Transformation

[Oxymoron]

Homework Statement

A random variable [tex]X[/tex] is distributed uniformly on [tex][-1,1][/tex]. Find the distribution of [tex]X^2[/tex], its mean and variance.

The Attempt at a Solution

Define a transformation of random variable as [tex]Y=X^2[/tex]. Problem is that the transformation function is not monotonic on the range. If it was just on [0,1] or [-1,0] then it would be monotonic and I could find inverses and hence define the cumulative distribution function, then differentiate to find the distribution.

Is this a correct deduction? Or is there more to it?

 

[micromass]

This is correct. So you can not apply the theorem directly because the square function is not increasing or decreasing on [-1,1].

There are two ways around this:
- Using a generalization of the theorem. Maybe you seen this:

Let X have pdf f_X, let Y=g(X). Let S be the sample space. Suppose there is a partition of A₀,...,A_k of S such that [tex]P(X\in A_0)=0[/tex] and that f_X is continuous on each A_i. Further, suppose that there exists functions g₁,...,g_k on A₁,..., A_k such that

• g(x)=g_i(x) for x in A_i.
• g_i(x) is monotone on A_i.
• The range Y=g_i(A_i) is the same for each i
• g_i^-1 has a continuous derivative on Y

Then [tex]f_Y(y)=\sum_{i=1}^k{f_X(g_i^{-1}(y))\left|\frac{d}{dy}g_i^{-1}(y)\right|}[/tex] for y in Y and 0 otherwise.

Apply this theorem with [tex]A_0=\emptyset,~A_1=[-1,0],~A_2=[0,1][/tex] and g(x)=x².

- If you haven't seen this theorem, then a direct computation of the distribution of [tex]Y=X^2[/tex] is also possible:

[tex]F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(-\sqrt{y}<X\leq \sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})[/tex]

Now differentiate it to get the pdf.

 

[Oxymoron]

Okay, so when I differentiate I get
[tex]f_Y(y) = \frac{\mbox{d}}{\mbox{d}y}F_Y(y) = \frac{\mbox{d}}{\mbox{d}y}F_X(\sqrt{y})\cdot\frac{1}{2\sqrt{y}}-\frac{\mbox{d}}{\mbox{d}y}F_X(-\sqrt{y})\cdot\frac{1}{-2\sqrt{y}}[/tex]
[tex]=f_X(\sqrt{y})\cdot\frac{1}{2\sqrt{y}}-f_X(-\sqrt{y})\cdot\frac{1}{-2\sqrt{y}}[/tex]
[tex]=\frac{1/2}{2\sqrt{y}}-\frac{1/2}{-2\sqrt{y}}[/tex]
[tex]=\frac{1}{2\sqrt{y}}[/tex]

and this is a PDF since

[tex]\int_{0}^{1}\frac{1}{2\sqrt{y}}\mbox{d}y = 1[/tex]

I can change the limits of the integral because I have already imposed that this integral exists on the partition [tex]0\leq y < 1[/tex].

 

[Oxymoron]

I also tried using the theorem and got the same answer.

What about the expected value?

By definition:

[tex]E[Y] = \int_{-\infty}^{\infty}y\frac{1}{2\sqrt{y}}\mbox{d}y = \infty[/tex]

But can I change the limits to 0 and 1 so that

[tex]E[Y] = \int_0^1y\frac{1}{2\sqrt{y}}\mbox{d}y = \frac{1}{3}[/tex]

is this something I can do? Or is the mean (and hence the variance) actually infinity for [tex]Y[/tex]?

 

[micromass]

Okay, so when I differentiate I get
[tex]f_Y(y) = \frac{\mbox{d}}{\mbox{d}y}F_Y(y) = \frac{\mbox{d}}{\mbox{d}y}F_X(\sqrt{y})\cdot\frac{1}{2\sqrt{y}}-\frac{\mbox{d}}{\mbox{d}y}F_X(-\sqrt{y})\cdot\frac{1}{-2\sqrt{y}}[/tex]
[tex]=f_X(\sqrt{y})\cdot\frac{1}{2\sqrt{y}}-f_X(-\sqrt{y})\cdot\frac{1}{-2\sqrt{y}}[/tex]
[tex]=\frac{1/2}{2\sqrt{y}}-\frac{1/2}{-2\sqrt{y}}[/tex]
[tex]=\frac{1}{2\sqrt{y}}[/tex]

This is not entirely correct, you have said [tex]f_X(\sqrt{y})=1/2[/tex], but this is only true if [tex]y\leq 1[/tex]. So for y>1, you would have [tex]f_Y(y)=0[/tex]. So your density would actually be [tex]f_Y(y)=\frac{1}{2\sqrt{y}}[/tex] if [tex]y\in [0,1] and 0 otherwise.

This fixes your paradox with the calculation of the mean.

 

[Oxymoron]

Yes it does. Thank you micromass, I should be able to go on and calculate the expected value and variance of this pdf.

 

[micromass]

Also note that you don't necessairily need to know the pdf to calculate the variance and the mean, since you have

[tex]E[X^2]=\int{x^2f_X(x)dx},~Var(X^2)=E[X^4]-E[X^2]^2=\int{x^4f_X(x)dx}-\left(\int{x^2f_X(x)}\right)^2[/tex]

Here we have essentially used the more general formula

[tex]E[g(X)]=\int{g(x)f_X(x)dx}[/tex]

which is often called "the law of the lazy statistician"