# If $Y\sim\mathcal N_{X,\:\sigma^2}$ and $Z\sim\mathcal N_{Y,\:\sigma^2}$, can we show that $Z\sim\mathcal N_{X,\:2\sigma^2}$?

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?

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}

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