# Sum of two dependent normally distributed variables

I have a question involving the elegant result of the distribution of the sum of independent normal variables, when using the mgf.

My question is as follows. Let ${X}^T = (X_1,X_2)$ be bivariate normal distributed with mean ${\mu} = ( \mu_1,\mu_2)$ and covariance matrix $\Sigma = \begin{bmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} = \begin{bmatrix} \sigma_1^2 & \rho \sigma_1 \sigma_2\\ \rho \sigma_1 \sigma_2 & \sigma_2^2 \end{bmatrix}$

Furter let $Y = X_1 + X_2$, where $X_1 \sim N(\mu_1,\sigma_1^2)$ and $X_2 \sim N(\mu_2,\sigma_2^2)$. If $X_1$ and $X_2$ where independent then the mgf of $Y$ is \begin{align} M_Y(t) =& M_{X_1}(t)M_{X_2}(t) \nonumber \\ =&\exp \Big(t\mu_1 + \frac{t^2}{2}\sigma_1^2\Big)\exp \Big(t\mu_2 + \frac{t^2}{2}\sigma_2^2\Big) \nonumber \\ =&\exp(t\mu_1)\exp(t\mu_2)\exp\Big(\frac{t^2}{2}\sigma_1^2\Big)\exp\Big(\frac{t^2}{2}\sigma_2^2\Big) \nonumber \\ =&\exp\Big(t(\mu_1 + \mu_2)+\frac{t^2}{2}(\sigma_1^2+\sigma_2^2)\Big) \label{mgf_sum} \end{align} which means $$X_1 + X_2 \sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2)$$ I know that the desired result, given that the variables are not independent, should be $$X_1 + X_2 \sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2+2cov(X_1,X_2))$$

Is it a way to show the correct variance of the distribution of $X_1+X_2$ along the lines of the mgf derivation as above?

• You need to find mgf of bivariate normal distribution $M_{X_1,X_2}(t_1,t_2)=E\left[e^{t_1X_1+t_2X_2}\right]$ by integrating joint pdf. After that take $t_1=t_2$. – NCh Apr 26 '17 at 1:48
• Thanks for your hint @NCh. I can update the question with the way I have sketched the derivation of the theoretical distribution of $X_1 + X_2$ – user1571823 Apr 26 '17 at 19:53