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 $f_{Y|X}$?
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$\begingroup$ If $N$ is independent of $X$ then $(X,Y)=(X,X+N)$ has a bivariate normal distribution. $\endgroup$– Michael HardyAug 19, 2020 at 5:34
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$\begingroup$ Right. So I can assume some value $\rho$ as correlation coefficient and use the bivaraiate normal formula to represent $f_{y|x}$, right? $\endgroup$– BikasAug 19, 2020 at 5:36
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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.
answered Aug 19, 2020 at 9:30
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$\begingroup$ Does it mean that the mean of $Y|X$ is $\mu_X$ and variance is $\sigma_N^2$? $\endgroup$– BikasAug 19, 2020 at 14:56
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$\begingroup$ The mean of $Y|X$ is $X$ (not $\mu_X$) and variance is $\sigma_N^2$. The mean of $Y$ is however $$\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y \:|\: X]] = \mathbb{E}[X]=\mu_X$$ $\endgroup$ Aug 19, 2020 at 17:27
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$\begingroup$ So, if I use $X = x_1$, then the mean of $Y|X$ would be $x_1$. Correct? $\endgroup$– BikasAug 19, 2020 at 17:32
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1$\begingroup$ 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 $x_1$". $\endgroup$ Aug 19, 2020 at 17:35
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