Question

@hartnn stuck on part two of that problem now.. How would I find the probability density function of the normal random variable X=3A-B

Answer 1

Oh and I'm given that two standard normal variables A and B are independent.

Answer 2

\(
f_{U+V}(x) = \int_{-\infty}^\infty f_U(y) f_V(x - y)\,dy
= \left( f_{U} * f_{V} \right) (x)
\)

Answer 3

do you have options/choices?

Answer 4

Nope

Answer 5

hmm...my first thought was to just put X=3A-B in the definition of std, normal fn, but then i saw this...

Answer 6

trying to find something in the book, but as usual, they never assign probability problems that are similar to the examples solved in the book...

Answer 7

I think I have something here... So:
PDF:
\[f(x)\ge0 \space for \space all \space x \in \mathbb{R} \]

Answer 8

and
\[\int_{-\infty}^{\infty}f(x)dx=1\]

Answer 9

those r true for any pdf

Answer 10

We know that:
\[\int_{-\infty}^{\infty}x^2*f(x)dx=E[X^2]=1\]eh nvm, this doesn't really help

Answer 11

yeah, mean variance has not much to do here..

Answer 12

http://math.stackexchange.com/questions/77873/how-calculate-the-probability-density-function-of-z-x-1-x-2

Answer 13

Ok so the standard normal variable denoted by Z is the random variable with pdf:
\[f(x)=\frac{1}{\sqrt{2\pi}}*e^{-\frac{x^2}{2}}\]