Question

If x~n(3, 4, y~n(-2, 6, z~n(1, 3, and x and y and z are independent, what are the distribution, expectation and variance of s = 3x-2y z?

Answer 1

I assume
`T\sim\mathcal N(\mu,\sigma^2)` refers to a random variable
`T` following a
`\mathcal N`ormal distribution with mean
`\mu` and variance
`\sigma^2`.

Recall that the moment generating function for a random variable
`T\sim\mathcal N(\mu,\sigma^2)` is


`M_T(t)=\exp\left(\mu t+\frac12\sigma^2t^2\right)`

and that the moment generating function for a linear combination of random variables is equivalent to the product of each individual random variables moment generating functions; that is, if
`T=\displaystyle\sum_ia_iT_i`, then


`M_T(t)=\displaystyle\prod_iM_(T_i)(a_it)`

So you have


`M_S(t)=M_(3X-2Y+Z)(t)=M_X(3t)* M_Y(-2t)* M_Z(t)`

with


`M_X(3t)=\exp\left(3(3t)+\frac12(4)(3t)^2\right)=\exp\left(9t+\frac12(36)t^2\right)`

`M_Y(-2t)=\exp\left(-2(-2t)+\frac12(6)(-2t)^2\right)=\exp\left(4t+\frac12(24)t^2\right)`

`M_Z(t)=\exp\left(t+\frac12(3)t^2\right)`

which means


`M_S(t)=\exp\left((9+4+1)t+\frac12(36+24+3)t^2\right)=\exp\left(14t+\frac12(63)t^2\right)`

This is the moment generating function of yet another normal distribution, so that
`S\sim\mathcal N(14,63)`, with expectation 14 and variance 63.