- #1
mcfc
- 17
- 0
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
[tex]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.[/tex]
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 [tex]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.[/tex]
Is that correct?
A2-For the pdf of Y: [tex]G(Y) = P(Y \leq y) = P(2 - x \leq y) = P(x \geq 2-y)
[/tex] but I'm not sure how to proceed??
Thanks
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
[tex]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.[/tex]
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 [tex]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.[/tex]
Is that correct?
A2-For the pdf of Y: [tex]G(Y) = P(Y \leq y) = P(2 - x \leq y) = P(x \geq 2-y)
[/tex] but I'm not sure how to proceed??
Thanks