# PDF for sum of two Gaussian random vectors with noise

Let $$\mathcal{N}(\mathbf{x}; \boldsymbol{\mu}, \mathsf{\Sigma})$$ be a multivariate Gaussian Probability Density Function (PDF), so

$$$$\mathcal{N}(\mathbf{x};\boldsymbol{\mu},\mathsf{\Sigma}) := \frac{1}{\sqrt{\det2\pi\mathsf{\Sigma}}} \exp\Big(-\frac{1}{2}(\mathbf{x} - \boldsymbol{\mu})^{\top}\mathsf{\Sigma}^{-1}(\mathbf{x} - \boldsymbol{\mu})\Big),$$$$

where $$\mathbf{x}, \boldsymbol{\mu} \in \mathbb{R}^m$$ and $$\mathsf{\Sigma} \in \mathbb{R}^{m,m}$$ is Symmetric Positive Definite (SPD). Let $$\mathbf{Y}_1$$ and $$\mathbf{Y}_2$$ be two random vectors with the following joint PDF

$$\begin{equation*} p_{\mathbf{Y}_{1}, \mathbf{Y}_{2}}(\mathbf{y}_{1}, \mathbf{y}_{2}) = \mathcal{N} \left( \begin{bmatrix} \mathbf{y}_1 \\ \mathbf{y}_2 \end{bmatrix} ; \begin{bmatrix} \mathbf{0}_n \\ \mathbf{0}_m \end{bmatrix} , \begin{bmatrix} \mathsf{\Sigma}_{11} & \mathsf{\Sigma}_{12} \\ \mathsf{\Sigma}_{21} & \mathsf{\Sigma}_{22} \end{bmatrix} \right) \cdot \end{equation*}$$

Furthermore, let $$\boldsymbol{\mathcal{E}}$$ be a Gaussian noise vector with PDF $$p_{\boldsymbol{\mathcal{E}}}(\boldsymbol{\epsilon})=\mathcal{N}(\boldsymbol{\epsilon}; \mathbf{0}, \sigma_{\epsilon}^2\mathsf{I_n})$$, which is independent of $$\mathbf{Y}_1$$ and $$\mathbf{Y}_2$$. Define $$\mathbf{Z}_1=\mathbf{Y}_1+\boldsymbol{\mathcal{E}}$$. The question is to find the joint PDF $$p_{\mathbf{Z}_{1}, \mathbf{Y}_{2}}(\mathbf{z}_{1}, \mathbf{y}_{2})$$. I have been told that the answer is

$$\begin{equation*} p_{\mathbf{Z}_{1}, \mathbf{Y}_{2}}(\mathbf{z}_{1}, \mathbf{y}_{2}) = \mathcal{N} \left( \begin{bmatrix} \mathbf{z}_1 \\ \mathbf{y}_2 \end{bmatrix} ; \begin{bmatrix} \mathbf{0}_n \\ \mathbf{0}_m \end{bmatrix} , \begin{bmatrix} \mathsf{\Sigma}_{11}+\sigma_{\epsilon}^2\mathsf{I} & \mathsf{\Sigma}_{12} \\ \mathsf{\Sigma}_{21} & \mathsf{\Sigma}_{22} \end{bmatrix} \right)\cdot \end{equation*}$$

My Thoughts

By the marginalization property of Gaussians, we can conclude $$p_{\mathbf{Y}_1}(\mathbf{y}_1)=\mathcal{N}(\mathbf{y}_1; \mathbf{0}_n,\mathsf{\Sigma}_{11})$$ and $$p_{\mathbf{Y}_2}(\mathbf{y}_2)=\mathcal{N}(\mathbf{y}_2; \mathbf{0}_m,\mathsf{\Sigma}_{22})$$. I know that by the convolution theorem, the PDF for the sum of two independent random vectors is the convolution of their PDF. Furthermore, I know that the convolution of two Gaussians is a Gaussian. This implies that $$\mathbf{Z}_1$$ is Gaussian. We can also obtain its mean by linearity of expectation

\begin{align*} \mathbb{E}(Z_{1,i}) &= \mathbb{E}(Y_{1,i}) + \mathbb{E}(\mathcal{E}_{1,i}) = 0 + 0 = 0, \end{align*}

and its covariance by bilinearity and symmetry of covariance as

\begin{align*} \text{cov}(Z_{1,i},Z_{1,j}) &= \text{cov}(Y_{1,i},Y_{1,j}) + \text{cov}(\mathcal{E}_{i},\mathcal{E}_{j}) + 2 \text{cov}(Y_{1,i},\mathcal{E}_{j}) \\ &= \text{cov}(Y_{1,i},Y_{1,j}) + \sigma_{\epsilon}^2\,\delta_{ij} + 0 \\ &= (\mathsf{\Sigma}_{11})_{ij} + \sigma_{\epsilon}^2\,\delta_{ij}, \end{align*}

where $$\delta_{ij}$$ is the Kronecker's delta. So, we have $$p_{\mathbf{Z}_1}(\mathbf{z}_1)=\mathcal{N}(\mathbf{z}_1; \mathbf{0}_n,\mathsf{\Sigma}_{11}+\sigma_{\epsilon}\mathsf{I}_n)$$. But I don't know how to calculate $$p_{\mathbf{Z}_{1}, \mathbf{Y}_{2}}(\mathbf{z}_{1}, \mathbf{y}_{2})$$.

• Is the noise vector independent only of $Y_1$? May 30, 2023 at 21:42
• @Snoop: mmmm, I think we can also assume that it is independent of both $\mathbf{Y}_1$ and $\mathbf{Y}_2$. Does this help? May 30, 2023 at 21:52
• What are the components of the covariance matrix of z1 and y2? You just need to put the pieces together.
– Eric
May 30, 2023 at 21:57
• @Eric: I didn't get you. Could you write an answer in detail? :) May 30, 2023 at 21:59

Suppose $$X:\Omega \to \mathbb{R}^n$$ is normally distributed with covariance matrix $$\Sigma$$ and $$\varepsilon:\Omega \to \mathbb{R}^m,m is normally distributed with covariance matrix $$\Gamma$$ and independent of $$X$$. Then, $$Y=X+(\varepsilon,0_{n-m})$$ is normally distributed with covariance matrix

$$\Sigma^*=\Sigma+\begin{bmatrix}\Gamma&0_{m\times n-m}\\ 0_{n-m\times m}&0_{n-m\times n-m}\end{bmatrix}=\Sigma +\Gamma_0$$

To see this, let $$\xi \in \mathbb{R}^n$$ and calculate the moment generating function of $$Y$$ as follows

\begin{aligned} M_{Y}(\xi)=E[e^{\xi^{\top}Y}]&=E[e^{\xi^{\top}X+\xi^{\top}(\varepsilon,0_{n-m})}]\\ &=E[e^{\xi^{\top} X}]E[e^{\xi_1\varepsilon_1 +...+\xi_m \varepsilon_m}]\\ &=E[e^{\xi^{\top} X}]E[e^{(\xi_1,...,\xi_m)^{\top} \varepsilon}]\\ &=e^{\xi^{\top}\Sigma \xi/2}e^{(\xi_1,...,\xi_m)^{\top}\Gamma (\xi_1,...,\xi_m)/2}\\ &=e^{\xi^{\top}\Sigma \xi/2}e^{\xi^{\top} \Gamma_0 \xi/2}\\ &=e^{\xi^{\top}(\Sigma + \Gamma_0)\xi/2}\\ &=e^{\xi^{\top}\Sigma^*\xi/2}\end{aligned},

where the formula for the moment generating function of a multivariate Gaussian is given here.

• @HoseinRahnama you're welcome May 31, 2023 at 0:28
• I was thinking the other day about this solution and was wondering if $(\varepsilon,0_{n-m})$ really justifies as a continuous random variable. I mean, a random variable which is constant cannot be continuous. right? Jun 1, 2023 at 9:17
• Why should this be a problem? Our aim is to prove that $Y$ is normally distributed with some specific parameters. @HoseinRahnama Jun 1, 2023 at 9:43
• @HoseinRahnama I understand your concern, but we can safely say that no distributional properties of $(\varepsilon,0_{n-m})$ are ever used in the argument. Only the simple fact that $\xi^{\top}(\varepsilon,0_{n-m})=(\xi_1,...,\xi_m)^{\top}\varepsilon$ is used. Jun 1, 2023 at 10:16
• Indeed it is not needed to characterize the distribution of such random variable here :). It is not a trivial object in my opinion so it may deserve its own question. @HoseinRahnama Jun 1, 2023 at 10:29

$$(Y_1,Y_2)$$ is a multivariate Gaussian with covariance $$\Sigma$$. $$\epsilon$$ is Gaussian with variance $$\sigma^2$$, so $$(\epsilon,0)$$ is Gaussian with covariance matrix being $$\sigma^2$$ in the top left and 0’s elsewhere. If $$A$$ and $$B$$ are independent,

\begin{align} \text{Cov} (A+B,A+B) &= \text{Cov}(A)+\text{Cov}(A,B)+\text{Cov}(B,A)+\text{Cov}(B,B) \\ &= \text{Cov}(A,A)+\text{Cov}(B,B) \end{align}

We have that the sum of independent multivariate Gaussians is multivariate Gaussian whose mean is the sum of the means (by linearity) and whose covariance is the sum of the covariance (from above), so the mean of $$(Z_1,Y_2)$$ is $$0$$ and it’s covariance is the sun of the covariances:

$$\begin{bmatrix} \mathsf{\Sigma}_{11}+\sigma^2\mathsf{I} & \mathsf{\Sigma}_{12} \\ \mathsf{\Sigma}_{21} & \mathsf{\Sigma}_{22} \end{bmatrix}$$

• Eric, it seems that $(\epsilon, 0)$ is not Gaussian as its covariance matrix is singular. There seems to be a bug in your argument. :) Indeed, $(\epsilon, 0)$ is not a continuous random vector but it is a mixed random vector and to do calculations with such random vectors we need dirac-delta and stuff like that. Jun 1, 2023 at 10:29
• Err, ok, but that doesn’t really matter to the calculation. If it makes you feel better, we can add $\epsilon_2$ to all the zero terms and then take the result as it approaches 0.
– Eric
Jun 1, 2023 at 12:00
• It would be good if you point this out in your answer to avoid confusion of future readers. :) Jun 1, 2023 at 12:11