Bivariate normal distribution from normal linear combination

[fisher garry]

I can't prove this proposition. I have however managed to prove that the linear combinations of the independent normal rv's are also normal by looking at it's mgf

$$E(e^{X_1+X_2+...+X_n})=E(e^{X_1})E(e^{X_2})...E(e^{X_n})$$
The mgf of a normal distribution is $$e^{\mu t}e^{\frac{t^2 \sigma^2}{2}}$$
$$E(e^{X_1+X_2+...+X_n})=e^{\mu_1 t}e^{\frac{t^2 \sigma_1^2}{2}}e^{\mu_2 t}e^{\frac{t^2 \sigma_2^2}{2}}...e^{\mu_n t}e^{\frac{t^2 \sigma_n^2}{2}}=e^{(\mu_1+\mu_2+...\mu_n) t}e^{\frac{t^2 (\sigma_1^2+\sigma_2^2+...+\sigma_n^2)}{2}}$$

Which is the normal distribution with mean $$\mu_1+\mu_2+...\mu_n$$ and variance $$\sigma_1^2+\sigma_2^2+...+\sigma_n^2$$

I know about the theory in section 5.4. Some of it is presented here
$$g(y_1,y_2)=f(x_1,x_2)|\frac{\partial(x_1,x_2)}{\partial(y_1,y_2)}|$$Can anyone show how the proof they refer to by section 5.4 and matrix theory goes?

 

[Ray Vickson]

View attachment 236979
I can't prove this proposition. I have however managed to prove that the linear combinations of the independent normal rv's are also normal by looking at it's mgf

$$E(e^{X_1+X_2+...+X_n})=E(e^{X_1})E(e^{X_2})...E(e^{X_n})$$
The mgf of a normal distribution is $$e^{\mu t}e^{\frac{t^2 \sigma^2}{2}}$$
$$E(e^{X_1+X_2+...+X_n})=e^{\mu_1 t}e^{\frac{t^2 \sigma_1^2}{2}}e^{\mu_2 t}e^{\frac{t^2 \sigma_2^2}{2}}...e^{\mu_n t}e^{\frac{t^2 \sigma_n^2}{2}}=e^{(\mu_1+\mu_2+...\mu_n) t}e^{\frac{t^2 (\sigma_1^2+\sigma_2^2+...+\sigma_n^2)}{2}}$$

Which is the normal distribution with mean $$\mu_1+\mu_2+...\mu_n$$ and variance $$\sigma_1^2+\sigma_2^2+...+\sigma_n^2$$

I know about the theory in section 5.4. Some of it is presented here
$$g(y_1,y_2)=f(x_1,x_2)|\frac{\partial(x_1,x_2)}{\partial(y_1,y_2)}|$$Can anyone show how the proof they refer to by section 5.4 and matrix theory goes?

I think you are looking at the wrong problem. The description implies that
$$
\begin{array}{rcl}
U&=& a_1 X_1 + a_2 X_2 + \cdots + a_n X_n \\
V&=& b_1 X_1 + b_2 X_2 + \cdots + b_n X_n
\end{array}
$$
Here the $a_i$ and $b_i$ are constants.

You can work out $EU, EV, \text{Var}(U), \text{Var}(V)$ and $\text{Cov}(U,V)$ in terms of the $a_i, b_i$ and the original $\mu, \sigma$ of the $X_i$. Thus, you have a formula for the means and variance-covariance matrix in terms of $a_i , b_i, i=1,2,\ldots, n$.

I think that what the question is asking for is the converse: given $\mu, \sigma$, the means and the variance-covariance matrix of $U,V$, it wants you to either find appropriate $a_i, b_i$ that will work (that is, give the right $U,V$), or at least to show that such $a_i, b_i$ exist, even if not easy to find. Using moment-generating functions (as you did above) should be very helpful for this purpose.

 

[Ray Vickson]

View attachment 236979
I can't prove this proposition. I have however managed to prove that the linear combinations of the independent normal rv's are also normal by looking at it's mgf

$$E(e^{X_1+X_2+...+X_n})=E(e^{X_1})E(e^{X_2})...E(e^{X_n})$$
The mgf of a normal distribution is $$e^{\mu t}e^{\frac{t^2 \sigma^2}{2}}$$
$$E(e^{X_1+X_2+...+X_n})=e^{\mu_1 t}e^{\frac{t^2 \sigma_1^2}{2}}e^{\mu_2 t}e^{\frac{t^2 \sigma_2^2}{2}}...e^{\mu_n t}e^{\frac{t^2 \sigma_n^2}{2}}=e^{(\mu_1+\mu_2+...\mu_n) t}e^{\frac{t^2 (\sigma_1^2+\sigma_2^2+...+\sigma_n^2)}{2}}$$

Which is the normal distribution with mean $$\mu_1+\mu_2+...\mu_n$$ and variance $$\sigma_1^2+\sigma_2^2+...+\sigma_n^2$$

I know about the theory in section 5.4. Some of it is presented here
$$g(y_1,y_2)=f(x_1,x_2)|\frac{\partial(x_1,x_2)}{\partial(y_1,y_2)}|$$Can anyone show how the proof they refer to by section 5.4 and matrix theory goes?

I won't give a lot of details, but will expand a bit on my previous post.

Presumably, if you are given the means $EU, EV$ and the variance-covariance matrix of the pair $(U,V)$, you are allowed to specify just how to make up $U,V$ in terms of some iid random variables $X_1, X_2, \ldots, X_n$. That is, we are allowed to choose an $n$ in an attempt to prove the result.

When we are given the above information about $U$ and $V$, we are given five items of data: two means, two variances, and one covariance. So, we will need at least five "coefficients" altogether.

I tried $U = a_1 X_1 + a_2 X_2$ and $V = b_1 X_1 + b_2 X_2 + b_3 X_3$. The means, variances and covariance of $(U,V)$ can be expressed algebraically in terms of the $a_i, b_j$ and the underlying $\mu, \sigma$ of the $X_k.$ Thus, we obtain five equations in the five variables $a_1, a_2, b_1, b_2,b_3$.

I managed to get a solution, thus proving the result asked for; however, it would possibly take many hours (perhaps days) of algebraic work to do it manually, so I saved myself a lot of grief by using the computer algebra package Maple to do the heavy lifting.

I don't see how matrix properties would be useful here, but maybe the person setting the problem had another method in mind.