- #1

- 634

- 1

[tex]

Y=3X+2

[/tex]

and the PDF of x is zero everywhere but:

[tex]f(x)=\frac{1}{4}e^{\frac{-x}{4}}, x>0[/tex]

I correctly got the mean like so:

[tex]\mu=E(h(x))=\int^{\infty}_{0} h(x)f(x)[/tex]

and evaluated it to be 10.

I am unsure as to how I go about getting the variance now. In previous problems I use the usual formula of:

[tex]\sigma=\int^{b}_{a} (x-\mu)^{2}f(x)[/tex]

or the shortcut of:

[tex]\sigma^{2}=E[X^{2}]-\mu^{2}[/tex]

However I'm unsure what to do in the case where the random variable is dependent on another with a given PDF function for continuous random variables.