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