Question

Let x1,x2 be two independent Normal random variables such that x1,x2 have means μ1,μ2 and variance σ1^2,σ2^2 respectively. Let Y = a1X1+a2X2, show that Y is a normal distribution and specify the mean an standard deviation for Y.

Answer 1

Please help and explain

Answer 2

Recall the moment generating function for a normally distributed random variable \(T\) with mean \(\mu\) and variance \(\sigma^2\).
\[\mathrm{M}_T(t)=\mathbb{E}(e^{Tt})=\exp\left(\mu t+\dfrac{1}{2}\sigma^2t^2\right)\]
Before I go on, are you familiar with the basic properties of MGFs?

Answer 3

For constants \(k_1,k_2\) and independent random variables \(T_1,T_2\), you have
\[\large\begin{align*}\mathrm{M}_{k_1T_1+k_2T_2}(t)&=\mathbb{E}(e^{(k_1T_1+k_2T_2)t})\\\\
&=\mathbb{E}(e^{k_1T_1t})\mathbb{E}(e^{k_2T_2t})\\\\
&=\mathbb{E}(e^{(k_1t)T_1})\mathbb{E}(e^{(k_2t)T_2})\\\\
&=\mathrm{M}_{T_1}(k_1t)\mathrm{M}_{T_2}(k_2t)
\end{align*}\]

Answer 4

So, if \(X_1\) is normally distributed with mean \(\mu_1\) and variance \({\sigma_1}^2\), then its MGF must be
\[\mathrm{M}_{X_1}(t)=\exp\left(\mu_1t+\frac{1}{2}{\sigma_1}^2t^2\right)\]
Similarly, for \(X_2\), you have
\[\mathrm{M}_{X_2}(t)=\exp\left(\mu_2t+\frac{1}{2}{\sigma_2}^2t^2\right)\]
From the property above, you have
\[\mathrm{M}_{a_1X_1}(t)=\mathrm{M}_{X_1}(a_1t)=\exp\left(\mu_1\color{red}{(a_1t)}+\frac{1}{2}{\sigma_1}^2\color{red}{(a_1t)}^2\right)\]
Try putting all this together.

Answer 5

okay cool so I set up the same format for a2 and X2

Answer 6

@sithsandgiggles

Answer 7

\[M _{a _{2}} X _{2} (t) = M _{X _{2}} (a _{2} t ) = \exp (\mu _{2} (a _{2} t) +\frac{ 1 }{ 2 } \sigma _{2} ^{2} (a _{2} t)^{2}\]

Answer 8

Right, and now the MGF of the sum of random variables, \(a_1X_1+a_2X_2\), is equal to the products of their respective MGFs, but their arguments are changed from \(t\) to \(a_1t\) and \(a_2t\), respectively.

In other words,
\[\mathrm{M}_{a_1X_1+a_2X_2}(t)=\mathrm{M}_{X_1}(a_1t)\times \mathrm{M}_{X_2}(a_2t)=\cdots\]