# Finding the distribution

1. Aug 22, 2009

### mcfc

Hello

I'm not too sure if this is the correct location for my post, but it's the best fit I can see!

The cdf of the continuous random variable X is
$$F(x)=\left\{\begin{array}{cc}0&\mbox{ if }x< 0\\ {1\over 4} x^2 & \mbox{ if } 0 \leq x \leq 2\\ 1 &\mbox{ if } x >2\end{array}\right.$$

Q1-Obtain the pdf of X
Q2-If Y = 2 - X, derive the pdf of the random variable Y

A1-I think the cdf is given by $$f(x) = F'(x)=\left\{\begin{array}{cc}{1\over 2}x &\mbox{ if } 0 \leq x \leq 2 \\ 0 &\mbox{ elsewhere } \end{array}\right.$$
Is that correct?

A2-For the pdf of Y: $$G(Y) = P(Y \leq y) = P(2 - x \leq y) = P(x \geq 2-y)$$ but I'm not sure how to proceed??
Thanks

2. Aug 22, 2009

### mathman

A1 - you're right.

A2 $$G(y) = P(Y \leq y) = P(2 - X \leq y) = P(X \geq 2-y) = 1-P(X < 2-y)=1-F(2-y)$$
G'(y)=F'(2-y) for the pdf

3. Aug 22, 2009

### mcfc

Thanks,
Q1-how would I evaluate:
$$G(y) = 1-F(2-y)\ \mbox{to give the cdf? Do I just substitute 2-y into }{1 \over 4} x^2 \mbox{ to give } G(Y) = 1 - {1 \over 4} (2-y)^2 = 4y - {1 \over 4}y^2\ for \ 0 \leq y \leq 2$$
$$\mbox{for the cdf of Y? and so for the pdf: } 4-{1 \over 2}y$$

Q2-Why are you able to make this step: $$1-P(X < 2-y)=1-F(2-y)$$