How to Derive the PDF of a Continuous Random Variable?

[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&amp;\mbox{ if }x&lt; 0\\<br /> {1\over 4} x^2 &amp; \mbox{ if } 0 \leq x \leq 2\\<br /> 1 &amp;\mbox{ if } x &gt;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&#039;(x)=\left\{\begin{array}{cc}{1\over 2}x &amp;\mbox{ if } 0 \leq x \leq 2 \\<br /> 0 &amp;\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)<br /> but I'm not sure how to proceed??
Thanks

 

[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 &lt; 2-y)=1-F(2-y)
G'(y)=F'(2-y) for the pdf

 

[mcfc]

A1 - you're right.

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

Thanks,
Q1-how would I evaluate:
G(y) = 1-F(2-y)\<br /> \mbox{to give the cdf? Do I just substitute 2-y into }{1 \over 4} x^2 \mbox{ to give }<br /> 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 &lt; 2-y)=1-F(2-y)