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Let $\sigma>0$ and note that $$(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\mapsto\mathcal N_{x,\:\sigma^2}(B)\;\;\;\text{for }$$ is a Markov kernel on $\left(\mathbb R,\mathcal B(\mathbb R)\right)$. Now let $X,Y,Z$ be real-valued random variables on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname P\left[Y\in B\mid X\right]=\mathcal N_{X,\:\sigma^2}(B)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R)\tag1$$ and $$\operatorname P\left[Z\in B\mid Y\right]=\mathcal N_{Y,\:\sigma^2}(B)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R)\tag2$$ and $$\operatorname P\left[Z\in B\mid X\right]=\mathcal N_{X,\:2\sigma^2}(B)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R)\tag3?$$ We know that the sum of independent normally distributed random variables is again normally distributed with accumulated mean and variance. Can we use this here somehow?

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Let me attempt this.

Write: $$ Y = U_1 + \alpha (X - E(X)) + X \\ $$ where $U_1 \sim N(0, \sigma_1^2)$ and $X$, $U_1$ are not correlated.

Because $Y \sim N(.,.)$, if $X$ is not normally distributed, then $\alpha=0$ and $U_1 \sim N(0, \sigma^2)$.

But if $X$ is normally distributed, it is still true that $\alpha=0$ and $U_1 \sim N(0, \sigma^2)$. That is because if $\alpha \ne 0$, then $var(Y|X) \ne \sigma^2$, contradicting $P[Y∈B∣X]=N(X,\sigma^2)$.

Similarly, we can let $Z = U_2 + \beta (Y -E(Y)) + Y$, where $ U_2 \sim N(0,\sigma_2^2)$ and show $\beta=0$.

So: $$ Y = U_1 + X \\ Z = U_2 + Y $$

Let $U = U_1 + U_2$, then $U \sim N(0, 2\sigma^2)$.

This can further be written as: $$ \begin{aligned} Z &= (U_2 + U_1) + X \\ &= U + X \\ & \sim N(0, 2\sigma^2) + X \\ & \sim N(X, 2\sigma^2) \end{aligned} $$

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  • $\begingroup$ Thank you for your answer. The crucial point is the independence of $U_1$ and $U_2$. How do you prove that? $\endgroup$ – 0xbadf00d Jul 31 '19 at 15:11
  • $\begingroup$ edited my answer to show this. $\endgroup$ – Tom Bennett Jul 31 '19 at 15:57
  • $\begingroup$ @TomBennett $Z = U_2 \pmb{+ +} \beta (Y -E(Y) )+ Y$ would not be $Z = U_2 \pmb{+} \beta (Y -E(Y) )+ Y$? $\endgroup$ – manooooh Aug 1 '19 at 1:47
  • $\begingroup$ sorry. typo and fixed. I obviously can't type on my phone. $\endgroup$ – Tom Bennett Aug 1 '19 at 4:43
  • $\begingroup$ further clarified $\endgroup$ – Tom Bennett Aug 2 '19 at 15:14

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