# How can we show that if $X_i\sim\mathcal N_{\sigma^2}(X_{i-1},\;\cdot\;)$, then $X_2-X_0\sim\mathcal N_{2\sigma^2}(0,\;\cdot\;)$?

Let $$\sigma>0$$ and $$\mathcal N_{\sigma^2}$$ denote the normal distribution kernel. If $$X_i$$ is a real-valued random variable and $$(X_{i-1},X_i)\sim\mathcal L(X_i)\otimes\mathcal N_{\sigma^2}$$, i.e. $$\operatorname P\left[X_i\in\;\cdot\;\mid X_{i-1}\right]=\mathcal N_{\sigma^2}(X_i,\;\cdot\;)$$, then $$(X_{i-1},X_i-X_{i-1})=\mathcal L(X_{i-1})\otimes\mathcal N_{\sigma^2}(0,\;\cdot\;)$$. In particular, $$\xi_i:=X_i-X_{i-1}\sim\mathcal N_{\sigma^2}(0,\;\cdot\;)$$ and $$X_{i-1}$$ and $$\xi$$ are independent.

How can we conclude that $$\xi_1+\xi_2\sim\mathcal N_{2\sigma^2}(0,\;\cdot\;)$$?

Remark: What I actually want to do is the following: Starting with an initial value $$x_0$$, I'm consecutively sampling $$x_i\sim\mathcal N_{\sigma^2}(x_{i-1},\;\cdot\;)$$. If the conclusion in the question would be correct, I could compute $$x_n$$ from $$x_0$$ by sampling $$x_n\sim\mathcal N_{n\sigma^2}(x_0,\;\cdot\;)$$ instead of applying all the intermediate sampling steps.

Can we show the desired result in general or am I missing an independence assumption which is implicit in the described sampling scheme?

• Do you mean that $X_i\mid X_{i-1}=x\sim \mathcal{N}(x,\sigma^2)$? Then how is $X_0$ distributed?
– user140541
Nov 20, 2019 at 10:06
• @d.k.o. Yes, $\operatorname P\left[X_i\in\;\cdot\;\mid X_{i-1}=x\right]\mathcal N_{\sigma^2}(x,\;\cdot\;)$ for $\mathcal L(X_{i-1}$-a.a. $x$ immediately follows from $\operatorname P\left[X_i\in\;\cdot\;\right]=\mathcal N_{\sigma^2}(X_i,\;\cdot\;)$. $X_0$ may be arbitrary distributed. Nov 20, 2019 at 10:58
• It is not essential, but I am not sure if your use of the tensor product ($\otimes$) is correct. For example, $(X_{i-1},X_i)\sim\mathcal L(X_i)\otimes\mathcal N_{\sigma^2}$ suggests for me that $X_{i-1}$ has the law $\mathcal L(X_i)$ and that $X_i$ has the law $\mathcal N_{\sigma^2}$. (But I could be wrong...) Also, the probability laws of two random variables only factors like that, if they are independent. But $X_{i-1}$ and $X_i$ are not independent, hence their joint probability distribution can not be written as $\mathbb P_{X_{i-1}} \otimes \mathbb P_{X_i}$ Jan 10, 2020 at 15:46

In short: You are talking about a Gaussian random walk, and the core proof is here. If you only need to sample $$X_n$$, then you can skip the intermediate steps.

On the sum of independent normal distributed random variables: As you asserted already, stochastic independence is critical. If independence is missing, there is an counter-example: If $$X \sim \mathcal N(0,\sigma^2)$$ and $$Y \sim \mathcal N(0,\sigma^2)$$, then it could happen that $$X = -Y$$, which leads to $$X+Y=0$$.

However, if $$X$$ and $$Y$$ are independent, then the sum fulfills $$X + Y \sim \mathcal N(0,2 \sigma^2)$$; see here.

Therefore, if $$\xi_1$$ and $$\xi_2$$ are independent, then the claim from your question is true.

About your remark: Since the process you describe is a stochastic process and also a discrete martingal, it is helpful to consider it as a martingale transformation. However, you don't need to know what it is to understand the rest.

Let $$\xi_1,\xi_2,\dots$$ be stochastical independent samples of a normal distributed random variable $$\mathcal N(0,\sigma^2)$$.

For a fixed initial point $$x_0 \in \mathbb R^n$$, your iteration is equivalent to $$x_n = x_{n-1} + \xi_{n}, \quad \text{for all } n = 1,2,\dots$$

We could also write $$x_n = x_0 + \sum_{i=1}^n \xi_i.$$

If you want to translate this into a theoretical object, we would consider $$n$$ independent random variables $$\Xi_i$$ which are all identical distributed $$\Xi_i \sim \mathcal N(0,\sigma^2)$$.

Now we can transform the sequence of random increments $$(\Xi_i)_{i=1,\dots,n}$$ into a random variable which describes the $$n$$th iterate $$X_n = x_0 + \sum_{i=1}^n \Xi_i.$$ (It is often useful to construct a stochastic process by transforming a simpler one, here we transform $$(\Xi_i)_{i\in \mathbb N}$$ into $$(X_i)_{i\in \mathbb N}$$, which are both martingales if w.l.o.g. $$x_0=0$$.)

You already pointed out the essential independence, i.e. $$X_{n-1}$$ and $$\xi_n$$ are independent! This allows us to conclude by induction that $$X_1 = x_0 + \xi_1 \sim \mathcal N(x_0,\sigma^2),\\ X_2 = X_1 + \xi_2 \sim \mathcal N(x_0,2\sigma^2),\\ X_3 = X_2 + \xi_3 \sim \mathcal N(x_0,3\sigma^2),\\ \dots$$

For each step we use the proof about the sum of independent normal distributed random variables. We obtain your claim $$X_n \sim \mathcal N(x_0,n \sigma^2)$$.

Note: My answer has a different notation than yours since I am not used to work directly with the kernel of the normal distribution.