# linear combination and univariate normal

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!

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.

• 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. Oct 14, 2019 at 13:58
• 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.
– whuber
Oct 14, 2019 at 14:07
• 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 Oct 14, 2019 at 14:11
• 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.
– whuber
Oct 14, 2019 at 14:22
• I appreciate your kind help. Have a nice day! Oct 14, 2019 at 14:58