1. ## 2 more questions

1) Ley Y be a continuous random variable pdf f(y) = 1 - abs(y) for abs(y) less than or equal to 1. Set X = Y^2. Give the range of X. Derive the cdf and pdf of X.

2) Let X be a continuous random variable with pdf f(x) = x/2 when 0 < x < 2. Let Y = 1/X. Find the pdf of Y.

They are similar problems probably requiring the same method, but I don't understand what I'm supposed to do.

2. Originally Posted by Janu42
1) Ley Y be a continuous random variable pdf f(y) = 1 - abs(y) for abs(y) less than or equal to 1. Set X = Y^2. Give the range of X. Derive the cdf and pdf of X.

[snip]
There are a variety of approaches. Here is one:

cdf = F(x) = 0 for x < 0.

$cdf = F(x) = \Pr(X < x) = \Pr(Y^2 < x) = \Pr(-\sqrt{x} < Y < + \sqrt{x}) = \int_{-\sqrt{x}}^{+\sqrt{x}} 1 - |y| \, dy$

$= \int_{-\sqrt{x}}^0 1 + y \, dy + \int^{+\sqrt{x}}_0 1 - y \, dy$ for $0 \leq x \leq 1$.

cdf = F(x) = 1 for x > 1.

Note that the pdf is found from dF/dx.

Your other question can be done in a similar way.

If you need more help please post all your working and say where yuo get stuck.