Question

Let X be a continuous random variable with density functionf(x)={1−|x|0 for −1

Answer 1

Answer:

$f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1$

Step-by-step explanation:

For this case we want to find the density function for $Y=X^2$

And we have the following density function for the random variable X:

$f(X) = 1- |X|,-1 \leq X \leq 1$

So we can do this replacing $Y=X^2$

$F_Y (Y \leq y) = P(X^2 \leq y)$

If we apply square root on both sides we got:

$P(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_{-\sqrt{y}}^0 1+t dt +\int_{0}^{\sqrt{y}} 1-t dt$

And if we integrate we got this:

$F_Y (y) = [t+ \frac{t^2}{2}] \Big|_{-\sqrt{y}}^0+ [t -\frac{t^2}{2}] \Big|_{0}^{\sqrt{y}}$

And replacing we got:

$F_Y (y) = [0 -(-\sqrt{y} +\frac{y}{2})] + [\sqrt{y} -\frac{y}{2}]$

$F_Y (y) = 2 |\sqrt{y}| -y$

And if we want to find the density function we just need to derivate the pdf like this:

$f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1$