Distribution of $pX+(1-p) Y$ We have two independent, normally distributed RV's: $$X \sim N(\mu_1,\sigma^2_1), \quad Y \sim N(\mu_2,\sigma^2_2)$$ and we're interested in the distribution of $pX+(1-p) Y, \space p \in (0,1)$. 
I've tried to solve this via moment generating functions. Since $$X \perp Y \Rightarrow \Psi_X(t) \Psi_Y(t)$$ where for $ N(\mu,\sigma^2)$ we'll have the MGF
$$\Psi(t) = \exp\{ \mu t + \frac12 \sigma^2 t^2 \}$$
After computation I've got the joint MGF as 
$$\Psi_{pX+(1-p) Y}(t) = \exp\{ t(p \mu_1 +(1-p)\mu_2) + \frac{t^2}{2}(p^2 \sigma_1^2 +(1-p)^2 \sigma_2^2) \}$$
which would mean $$pX+(1-p) Y \sim N(p \mu_1 +(1-p)\mu_2, p^2 \sigma_1^2 +(1-p)^2 \sigma_2^2) $$
Is my approach correct? Intuitively it makes sense and the math also adds up.

Any linear combination $aX+bY$ of independent normally distributed random variables $X$ and $Y,$ where $a$ and $b$ are constants, i.e. not random, is normally distributed. You can show that by using moment-generating functions provided you have a theorem that says only normally distributed random variables can have the same m.g.f. that a normally distributed random variable has.
Alternatively, show that $aX$ and $bY$ are normally distributed, and then compute the convolution of the two normal density functions, to see that it is also a normal density function.
And basic formulas concerning the mean and the variance say
\begin{align}
\operatorname{E}(aX+bY) & = a\operatorname{E}(X) + b\operatorname{E}(Y), \\[6pt]
\operatorname{var}(aX+bY) & = a^2 \operatorname{var}(X) + b^2\operatorname{var}(Y).
\end{align}
