Computation with a transformation of a random variable.

Question

The random variable $X$ is uniformly distributed on the interval $[-4,4]$. Compute $P(X^2 ≤ 9)$.

I tried to define $Y = X^2$ and compute the density of this transformed variable. The PDF of the uniform distribution is $f_x(x) = \frac{1}{8}$. What I got as a transformed density is $f_y(y) = \frac{1}{16\sqrt(y)}$.

However, if I calculate the integral of that transformed density from $0$ to $9$, I get a wrong result. My guess is that it has something to do with the fact that the uniform can take negative values.

Answer 1

Hint: $$ P(X^2<9)=P(-3<X<3)=\int_{-3}^{3}f(x)\,dx $$ where $f$ is the density function for $X$.

Now what you need is to find the correct density function for $X$ and then integrate it.

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Notes.

It is fairly easy the see by intuitions of "uniform distribution" that $$ P(-3<X<3)=\frac{\textrm{length of the interval }[-3,3]}{\textrm{length of the interval}[-4,4]}=\frac34 $$

Answer 2

Your transformed density should be $f_Y(y) = \dfrac{1}{8\sqrt{y}}$ for $0 \lt y \le 4$.

You may have forgotten to account for two different values of $X$ corresponding to each value of $Y$

You then get $$\int_0^9 f_Y(y) \, dy = \frac34$$