Squaring uniform/normal distribution and expectation

[rukawakaede]

Suppose $$X$$ is a uniformly distributed random variable on an interval $$[-a,a]$$ for some real $$a$$.
Let $$Y=X^2$$. Then what could you say about this distribution of $$Y$$? I have no idea how to think about this distribution.
Also how could we compute the expectation of $$Y$$? I know that $$E[X]=0$$ but what could I conclude about $$E[Y]=E[X^2]$$ and $$E[XY]=E[X^3]$$?
Is E[Y]=Var[X] since E[X]=0?

Similarly suppose X~N(0,1) be a standard normal random variable. What could we say about distribution of $$Y=X^2$$?

Hope someone could help solving my confusion.

 

[micromass]

Hi rukawakaede,

The distribution of a square can easily be calculated as follows:

$$F_Y(y)=P\{Y\leq y\}=P\{X^2\leq y\}=P\{-\sqrt{y}\leq X\leq \sqrt{y}\}=P\{X\leq \sqrt{y}\}-P\{X<-\sqrt{y}\}=F_X(\sqrt{y})-F_X(-\sqrt{y})$$

where in the last step we've used that the distribution is continuous. Now, to obtain the pdf, just differentiate both sides.

Now, to obtain the expectation, you can calculate this with the distribution function obtained above. But there's a simpler way. The so-called "law of the lazy statistician" gives us that

$$E(g(X))=\int_{-\infty}^{+\infty}{g(x)f_X(x)dx}$$

So, in particular

$$E(X^2)=\int_{-\infty}^{+\infty}{x^2f_X(x)dx}$$

So, to obtain the expactation of X², there is no need to know the distribution of X². Only know the distribution of X is enough!