# PDF of mean of two PDF of gaussian vectors

I have $$f_1(x,y)$$ and $$f_2(x,y)$$ are PDFs for two 2-dimensional gaussian vectors that have same diagonal on covariance matrix and same mean but different correlation $$\rho_1, \rho_2 \neq \pm 1$$: $$X_1 \sim \mathcal{N} \left( \begin{pmatrix} \mu_x \\ \mu_y \end{pmatrix} , \begin{pmatrix} \sigma^2_x & \rho_1 \sigma_x \sigma_y \\ \rho_1 \sigma_x \sigma_y & \sigma^2_y \end{pmatrix} \right)$$ $$X_2 \sim \mathcal{N} \left( \begin{pmatrix} \mu_x \\ \mu_y \end{pmatrix} , \begin{pmatrix} \sigma^2_x & \rho_2 \sigma_x \sigma_y \\ \rho_2 \sigma_x \sigma_y & \sigma^2_y \end{pmatrix} \right)$$ These variables are independent and introduced just for visual explanations of conditions.

Let's define function: $$g(x,y) := \dfrac{1}{2}(f_1(x,y) + f_2(x,y))$$

I need to understand if $$g(x,y)$$ is PDF of some Gaussian vector (bivariate normal).

If we write PDF of gaussian vector then we see that in the case of different correlation coefficients corressponding coefficients near $$x^2$$ and $$y^2$$ would vary $$f(x,y)=\frac{1}{2\pi\sigma_{x}\sigma_{y}\sqrt{1-\rho^2}}\exp\left[-\frac{1}{2(1-\rho^2)}\left(\frac{(x-\mu_{x})^{2}}{\sigma^{2}_{x}}+\frac{(y-\mu_{y})^{2}}{\sigma^{2}_{y}}-\frac{2\rho(x-\mu_{x})(y-\mu_{y})}{\sigma_{x}\sigma_{y}}\right)\right]$$ Also, it seems to me that we actually sum two ellipses with different rotation angle so the resulting object is not ellipse in general.

So my hypothesis is the resulting PDF is not PDF of some Gaussian vector but i have no idea how to show it formally.

EDIT 1:

I have tried to plot the resulting function for variables with same sigma equals to 1 and different correlations (0.1, 0.9) wolfram and get something that not similar to ellipsoid but i still don't know how to show it formally.

• how are $X_1$ and $X_2$ related? May 9, 2023 at 22:39
• @lmaosome they are not related at all, i have introduced them just for the explanation of conditions May 10, 2023 at 8:56
• I see. You are looking for the density of $0.5\mathcal N(\mu,\Sigma_1) + 0.5\mathcal N(\mu,\Sigma_2)$; this kind of distribution is called Gaussian Mixture Distribution May 11, 2023 at 9:45
• @lmaosome really thanks for the keyword! May 11, 2023 at 13:45

I have showed it using property of gaussian PDF that is log-concave.

If we take natural log of gaussian PDF $$h(x,y) := \log f(x,y) = -\log \left(2\pi \sigma_x \sigma_y \sqrt{1 - \rho^2}\right) - \dfrac{1}{2(1-\rho^2)}\left(\dfrac{x^2}{\sigma_x^2} - 2\rho\dfrac{xy}{\sigma_x \sigma_y} + \dfrac{y^2}{\sigma_y^2} \right)$$

and calculate it's Hessian matrix $$\mathcal{H}_h = \begin{pmatrix} - \dfrac{1}{(1-\rho^2)\sigma_x ^2} & \rho \dfrac{1}{\sigma_x \sigma_y (1-\rho^2)} \\ \rho \dfrac{1}{\sigma_x \sigma_y (1-\rho^2)} & -\dfrac{1}{(1-\rho^2)\sigma_y ^2} \end{pmatrix}$$ Then by Sylvester criterion it is negative definite: $$\mathcal{H}_h^{11} = - \dfrac{1}{(1-\rho^2)\sigma_x ^2} < 0$$ $$\det \mathcal{H}_h = \dfrac{1}{(1-\rho^2)^2\sigma_x ^2 \sigma_y^2} - \rho^2 \dfrac{1}{(1-\rho^2)^2\sigma_x ^2 \sigma_y^2} = \dfrac{1 - \rho^2}{(1-\rho^2)^2\sigma_x ^2 \sigma_y^2} > 0 \quad |\rho| < 1$$

so the $$h(x,y)$$ is concave and $$f(x,y)$$ is log-concave.

Then we could take $$\mu_x = \mu_y = 0 \quad \sigma_x = \sigma_y = 1$$ $$\rho_1 = - \rho_2 = \dfrac{4}{5}$$

And calculate values in points $$(-2, 2)$$, $$(2, 2)$$. $$f_1(-2,2) = f_2(2,2) = \dfrac{5}{6\pi} \exp {\left[ -20 \right]}$$

$$f_2(-2,2) = f_1(2,2) = \dfrac{5 }{6\pi} \exp {\left[ -\dfrac{20}{9} \right]}$$ and $$(0, 2) = \dfrac{1}{2}((-2, 2) + (2, 2))$$ $$f_1(0,2) = f_2(0,2) = g(0,2) = \dfrac{5}{6\pi} \exp {\left[ -\dfrac{50}{9} \right]}$$

Then from log-concave property next must holds $$g(0,2) = g\left((-2, 2) \dfrac{1}{2} + (2, 2) \dfrac{1}{2}\right) \geqslant g(-2, 2)^{\tfrac{1}{2}} \cdot g(2, 2)^{\tfrac{1}{2}} = g(2, 2)$$

but $$g(0,2) = \dfrac{5}{6\pi} \exp {\left[ -\dfrac{50}{9} \right]} < \dfrac{5}{6\pi} \exp {\left[ -\dfrac{50}{9} \right]} \left(\underbrace{\dfrac{\exp {\left[ \dfrac{-130}{9} \right]} + \exp {\left[ \dfrac{30}{9} \right]}}{2}}_{> 1}\right) = g(2,2)$$