# Regular Version of Conditional Gaussian Distribution

Let $Z_{1}$ and $Z_{2}$ be two independent normally distributed random variables with expectations $\mu_{1},\mu_{2}\in\mathbb{R}$ and variances $\sigma_{1}^2,\sigma_{2}^2\in (0,\infty)$ .

I would like to prove that the regular conditional distribution $\mathbb{P}(Z_{1}\in \cdot$ $|Z_{1}+Z_{2}=x)$, for almost every $x\in\mathbb{R}$, is equal to the normal distribution with expectation $\mu_{1}+\frac{\sigma_1^2}{\sigma_1^2+\sigma_2^2}(x-\mu_1-\mu_2)$ and variance $\frac{\sigma_2^2\sigma_1^2}{\sigma_1^2+\sigma_2^2}$.

I don't know how to start. Can anybody help, please?

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The beginning has to be, in one way or another, to write down the joint distribution of $(Z_1,Z_1+Z_2)$. Can you do that? – Did Dec 3 '12 at 9:30
I'm not sure if I do the right.. Do you mean, there is a density function $f_{Z_1,Z_1+Z_2}(z_1,x)$ so that holds $\int \int f_{Z_1,Z_1+Z_2}(z_1,x)dx$ $dz_1=1$ where $x=z_1+z_2$? – Ichigo Dec 3 '12 at 10:18
If you cancel "where $x=z_1+z_2$", then your comment is right. – Did Dec 3 '12 at 10:21
So what shall I do with the joint distribution? How can I combine this with the conditional distribution? – Ichigo Dec 3 '12 at 10:34
If you know $f_{Z_1,Z_1+Z_2}$ and $f_{Z_1+Z_2}$, you do have access to the conditional density of $Z_1$ conditionally on $Z_1+Z_2$, no? – Did Dec 3 '12 at 10:42