# Distribution after adding noise to a Gaussian Distribution

I have a random variable $$X \sim \mathcal{N}(\mu_X, \sigma_X^2)$$. Now, I add a noise $$N \sim \mathcal{N}(0, \sigma_N^2)$$ to $$X$$ to get $$Y$$ ($$Y = X + N$$). Thus, $$Y \sim \mathcal{N}(\mu_X, \sigma_X^2 + \sigma_N^2)$$. Are $$Y$$ and $$X$$ bivariate normal now? If so, what would be the density function $$f_{Y|X}$$?

• If $N$ is independent of $X$ then $(X,Y)=(X,X+N)$ has a bivariate normal distribution. Aug 19, 2020 at 5:34
• Right. So I can assume some value $\rho$ as correlation coefficient and use the bivaraiate normal formula to represent $f_{y|x}$, right? Aug 19, 2020 at 5:36

We know that $$X+N\: | \: X=x \sim x+N \: | \: X=x.$$ If we also know that $$X$$ and $$N$$ are independent, then we can infer that $$x+N \: | \: X=x \sim x+N \sim \mathcal{N}(x, \sigma_N^2),$$ which means that $$Y|X \sim \mathcal{N}(X,\sigma_N^2)$$.
For the part about bivariate normal distribution, you may use that linear transformations of a bivariate normal vector is again bivariate normal. So if $$(X,N)$$ is bivariate normal (which it is if we assume independence), then $$(X,X+N)$$ is also bivariate normal.
• Does it mean that the mean of $Y|X$ is $\mu_X$ and variance is $\sigma_N^2$? Aug 19, 2020 at 14:56
• The mean of $Y|X$ is $X$ (not $\mu_X$) and variance is $\sigma_N^2$. The mean of $Y$ is however $$\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y \:|\: X]] = \mathbb{E}[X]=\mu_X$$ Aug 19, 2020 at 17:27
• So, if I use $X = x_1$, then the mean of $Y|X$ would be $x_1$. Correct? Aug 19, 2020 at 17:32
• Yes, although when a particular value of $X$ is considered, you should write $Y|X=x_1$, as in "The mean of $Y|X=x_1$ is $x_1$". Aug 19, 2020 at 17:35