# Variance of Marginals of Continuous Random Walk Conditioned on Future Value

Consider the $$N$$ i.i.d. values

$$X_i \sim \mathcal{N}(0, \sigma^2)$$

such that

$$Z_i = \sum_{j=1}^i X_j$$

I am interested in the distribution

$$f(X_i | Z_N = z)$$

Mean

Under the condition, the $$X_i$$ are no longer independent, but they should remain identically distributed. Hence

$$Z_N = \sum_{i=1}^N X_i = z$$

implies

$$\sum_{i=1}^N \mathbb{E}[X_i] = z$$

and therefore

$$\mathbb{E}[X_i] = z/N$$

Variance

Empirically I found the variance to be

$$var(X_i) = \frac{N-1}{N} \sigma^2$$

Distribution

If you start doing the calculations to find the distribution working backwards from $$X_N$$ I believe you are simply finding the product of many Gaussian distributions so I believe the distribution is going to be Gaussian. However I ran into issues with this after the first step.

Question

How do I show rigorously that the marginal distribution

$$(X_i | Z_N = z) \sim \mathcal{N} \left (\frac{z}{N}, \frac{N-1}{N} \sigma^2 \right)$$

It all follows from the properties of multivariate normals. Since $$X_i$$ are independent and normally distributed, they're jointly normal, which means any linear combination of them is also jointly normal with them. So, $$p_{\mathbf{X},Z_N}(\mathbf{x},z)$$ is a multivariate normal, which in turn means $$p_{X_i,Z_N}(x,z)$$ is multivariate normal with

$$\mu=\begin{bmatrix}0\\0\end{bmatrix},\Sigma=\begin{bmatrix}\sigma^2&\sigma^2\\\sigma^2&N\sigma^2\end{bmatrix}$$

Because $$p_{X_i,Z_N}(x,z)$$ is MV normal, the conditional distributon $$p_{X_i|Z_N}(x,z)$$ is univariate normal, and conditional expectation and the variance can be found (same as your answer) via the formulas under "Conditional distributions" section in the wiki page linked above.

It's a bit messy, which might explain why it isn't seen more. Here is a sketch starting from the bivariate case, which generalizes. I'll use $$X$$ and $$Y$$ and $$Z=X+Y.$$

First let's find the conditional cdf for $$Z$$ given $$X=x.$$

$$F_{Z|X=x}=P \left[ X+Y \leq z \ | \ X=x\right]=P[Y \leq z-x]=F_Y(z-x)$$

Then the conditional pdf is found by differentiating: $$f_{Z|X=x}=f_{Z|X}(z|x)=f_Y(z-x)$$

That in turn means the joint density function is

$$f_{X,Z}(x,z) = f_{Z|X}(z|x)f_X(x)=f_Y(z-x)f_X(x)$$

Finally, we get the conditional density formula

$$f_{X|X+Y=z}(x,z)=\frac{f_Y(z-x)f_X(x)}{f_Z(z)} \ \ \ \ \ \ [1]$$

Now let's look at the normal case. Let $$Z=X_1 + X_2 + \cdots + X_N$$ and $$Y = X_1 + X_2 + \cdots + X_{N-1},$$ where

$$X_i \sim \mathcal{N} \left( \mu, \sigma^2 \right)$$ and $$Y \sim \mathcal{N} \left( \left( N - 1 \right) \mu, \left( N-1 \right) \sigma^2 \right)$$

Now what is the pdf $$f_{X_N|Z}(x,z)$$?

Using $$[1],$$

$$f_{X|Z=z}(x,z) = \frac{ \left( \frac{1} {\sqrt{{2 \pi \left(N-1 \right)\sigma^2} } } \right) \left( e^{\frac{- \left( z-x- \left( N-1 \right) \mu \right)^2 } {2 \left( N - 1 \right) \sigma^2}} \right) \left( \frac{1}{\sqrt{2 \pi \sigma^2}} \right) \left( e^{\frac{- \left( x- \mu \right)^2}{2 \sigma^2}} \right) } {\left( \frac{1} {\sqrt{{2 \pi N \sigma^2} } } \right) \left( e^{\frac{- \left( z - N \mu \right)^2}{2 N \sigma^2}} \right)}$$

After much simplifying (probably worth going throug once, but I'm not putting the details in here at this time), this can be expressed as

$$f_{X|Z=z}(x,z) =\left( \frac{1}{\sqrt{2 \pi s^2 }} \right) \left( e^{\frac{- \left( x - \frac{z}{N} \right)^2}{2 s^2}}\right),$$

where $$s^2 = \frac{\left( N - 1 \right) \sigma^2}{N}$$

This is now recognizable as a normal pdf and we can see that $$X_i|Z \sim \mathcal{N} \left( {\frac{z}{N}}, \frac{N-1}{N} \sigma^2 \right)$$ as you found empirically.