Consider the following bivariate normal distribution: $N((1,0), \textbf{I})$. When we sample $10$ means from this distribution, how exactly does this work? Is each mean a linear combination of $\mu_1 = 1$ and $\mu_2 = 0$?
Asked
Active
Viewed 61 times
1 Answers
0
Bivariate distributions produce a pair (vector) of values. Let's call the pair $(X,Y)$. Then in this case: $X\sim N(0,1)$ and $Y\sim N(1,1)$ and they are not correlated.
I'm assuming that by "sampling means" you are referring to the value of $M = (X+Y)/2$. If this is indeed what you mean, we can work out the distribution by convolving the two normals (standard addition of random variables).
$$2M = X+Y \implies p_{2M}(t) = (p_X*p_Y)(t) $$
$$p_{2M}(t)=\frac{1}{2\pi}\int_\infty^\infty \left(\exp{-\frac{(x-t)^2}{2}}\right) \left(\exp{-\frac{(x-1)^2}{2}}\right) dx$$
$$ = \frac{e^{-\frac{1}{4} (t-1)^2}}{2 \sqrt{\pi }}$$
Which we can see is a normal with mean of $1$ and variance of $2$, i.e. $2M\sim N(1,2)$. Rescaling from $2M$ to $M$, gives $M\sim N(1/2,1/2)$.
Applying the same process more generally for $X\sim N(a,s^2)$ and $Y\sim N(b,t^2)$ we get: $$ \begin{array}{c|c|c} \text{Variable} & \text{Mean} & \text{Variance} \\ \hline X & a & s^2 \\ \hline Y & b & t^2 \\ \hline X+Y & a+b & s^2 + t^2 \\ \hline \frac{X+Y}{2} & \frac{a+b}{2} & \frac{s^2 + t^2}{4} \end{array} $$
Lucas
- 1,469