The [[Independence of Random Variables]] for a [[Gaussian Distribution]] establishes the following [[Joint Probability Density Function|Joint PDF]]. **Standard Gaussian:** * $X \sim \mathcal N(0,1)$ * $Y \sim \mathcal N(0,1)$ * $X \perp Y$ $ \begin{align} f_{X,Y}(x,y) &=f_X(x)*f_Y(y) \tag{1}\\[6pt] &=\frac{1}{2 \pi}* \exp(-\frac{x^2}{2}) * \frac{1}{2 \pi}* \exp(-\frac{y^2}{2}) \tag{2}\\[8pt] &=\frac{1}{2 \pi}* \exp\big (-\frac{1}{2}*(x^2+ y^2)\big) \tag{3} \end{align} $ (3) Products of separate exponentials can be written as a single exponential with summed exponents. **General Gaussian:** * $X \sim \mathcal N(\mu_x,\sigma_x^2)$ * $Y \sim \mathcal N(\mu_y,\sigma_y^2)$ * $X \perp Y$ $ \begin{align} f_{X,Y}(x,y) =& f_X(x) * f_Y(y) \\ =&\Big(\frac{1}{\sigma_x \sqrt{2 \pi}}*\exp\big(-\frac{(x-\mu_x)^2}{2 \sigma_x^2}\big)\Big) * \\ &\Big(\frac{1}{\sigma_y \sqrt{2 \pi}}*\exp\big(-\frac{(y-\mu_y)^2}{2 \sigma_y^2}\big)\Big) \\ =&\frac{1}{2\sigma_x \sigma_y \pi} * \exp\Big(-\frac{(x-\mu_x)^2}{2\sigma_x^2} -\frac{(y-\mu_y)^2}{2\sigma_y^2} \Big) \end{align} $ The joint density has its maximum, where the exponent $=0$. This will be true, when $\{x,y\}$ are equal to $\{\mu_x, \mu_y\}$, which is also the expectation of the joint PDF. ![[independent-joint-gassuain.png|center|400]]