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Show that $(X_1,X_2)$ has a bivariate normal distribution with means $\mu_1, \mu_2$, variances $\sigma _1^2 $ and $\sigma _2^2$, and correlation coefficient $\rho $ if and only if every linear combination $t_1X_1+t_2X_2$ has a univariate normal distribution with mean $t_1\mu_1+t_2\mu_2$, and variances $t_1^2\sigma _1^2+t_2^2\sigma _2^2+2t_1t_2\rho \sigma_{12}$ where $t_1, t_2$ are non zero constants

My work:

Joint mgf of $X_1 $and $X_2=e^{[t_1\mu_1+t_2\mu_2+\frac{1}{2}(t_1^2\sigma _1^2+t_2^2\sigma _2^2+2\rho \sigma_1 \sigma_1t_1t_2 )]} $

I am confused here to find the proper grouping to get the forward result.

For the backward result, I know I have to find the m.g.f. of $t_1X + t_2Y$, for any $t_1 $and $t_2$ real,and plug in the $E(t_1X + t_2Y)$ and $Var(t_1X + t_2Y)$.

I am struggling to get the correct backward result.

Second part:

Let $Y_i=X_i/\sigma_i, i=1,2$ Show that $Var(Y_1-Y_2)=2(1-\rho)$

I have found, $Var(Y_1)=\sigma_1, Var(Y_2)=\sigma_2$ since $X_1/\sigma_1=\sigma _1$

How can I prove that $Var(Y_1-Y_2)=2(1-\rho)$? Thank you so much for your help!

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It's simpler than it looks.

To avoid writing lots of exponentials, let's work with the cumulant generating functions. These are the logarithms of the characteristic function: that is, the cgf of any random variable $X$ is

$$\psi_X(s) = \log E\left[e^{isX}\right].$$

For two variables $(X,Y)$ the cgf is

$$\psi_{X,Y}(s,t) = \log E\left[e^{isX+itY}\right].$$

The cgf of a Normal variable with mean $\mu$ and variance $\sigma^2$ is $i\mu s - \sigma^2 s^2/2.$ Thus you know there exist constants $\mu_1,$ $\mu_2,$ $\sigma_1,$ and $\sigma_2$ for which

$$\eqalign{\log E\left[e^{i u(t_1X_1+t_2X_2)}\right] &= \psi_{t_1 X_1 + t_2 X_2} (u)\\&= i(t_1\mu_1+t_2\mu_2)u - (\sigma_1^2t_1^2 + 2\rho \sigma_1\sigma_2 t_1 t_2 + \sigma_2^2 t_2^2) u^2/2}$$

no matter what the (real) values of $t_1, t_2,$ or $u$ might be.

But given any real numbers $s$ and $t,$ you can find numbers $u,$ $t_1,$ and $t_2$ (for instance, $u=1,$ $t_1=s,$ and $t_2=t$) for which $s = ut_1$ and $t=ut_2.$ It needs only basic rules of algebra to re-express the preceding result in terms of $s$ and $t:$

$$\psi_{X_1,X_2}(s,t) = \log E\left[e^{i s X_1+i t X_2}\right] = i(s\mu_1+t\mu_2) - (\sigma_1^2 s^2 + 2\rho \sigma_1\sigma_2 s t + \sigma_2^2 t^2)/2.$$

The right hand side is the cgf of the bivariate Normal distribution with mean $(\mu_1,\mu_2)$ , variances $\sigma_1^2$ and $\sigma_2^2,$ and correlation $\rho.$ Because the cgf uniquely determines the distribution (which is why we work with the cgf rather than the mgf!), we're done.

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  • $\begingroup$ I am kinda new in the statistics field. I know $logM_{x}= logE(e^{tx})$ but i am not familiar with $$\psi_X(s) = \log E\left[e^{isX}\right].$$ Could you please give any reference to know more about this formula. Also how can I prove the second part? Thnk you so much for your time. $\endgroup$ – Barsal Oct 14 '19 at 13:58
  • $\begingroup$ The second part is comparatively trivial: from the expression for the cgf of a bivariate normal it is immediate that all the linear combinations of its components are themselves normal. For references, see Wikipedia (online) or Stuart & Ord, Kendall's Advanced Theory of Statistics. You usually can work with the mgf if you like--the manipulations are exactly the same--but you ought to pay attention to questions of existence and uniqueness. $\endgroup$ – whuber Oct 14 '19 at 14:07
  • $\begingroup$ What about the forward results that is, how to show that $(X_1,X_2)$ has a bivariate normal distribution with means $\mu_1, \mu_2$, variances $\sigma _1^2 $ and $\sigma _2^2$, and correlation coefficient $\rho $ implies that every linear combination $t_1X_1+t_2X_2$ has a univariate normal distribution $\endgroup$ – Barsal Oct 14 '19 at 14:11
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    $\begingroup$ As I wrote, that is dead easy. The point of the cgf approach is that a quadratic form in $n$ (here, $n=2$) variables is quadratic in all their linear combinations and, conversely, a function that is quadratic in all linear combinations must be a quadratic form. Both of these are easy to prove using basic algebra, thereby taking out of the question any potentially complicated, advanced, or non-intuitive manipulations of distributions or random variables. $\endgroup$ – whuber Oct 14 '19 at 14:22
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    $\begingroup$ I appreciate your kind help. Have a nice day! $\endgroup$ – Barsal Oct 14 '19 at 14:58

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