- #1
Lancelot59
- 646
- 1
I have the continuous random variable Y, defined such that:
[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.
[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.