# Confusion regarding random walk model

I was referring to this book where it is given that

If we assume equally spaced nodes $i$ for $i=1,...,n$.

The first order random walk is constructed using independent increments

$$\Delta{x_i} \sim N(0,k^{-1}), \ \ \ \ \ \ i=1,...,n-1$$

The density for $${\bf{x}}=(x_1,x_2...x_n)'$$ is derived from the $n-1$ increments as

$$\pi({\bf{x}}|k) = k^{(n-1)/2} \cdot \exp \left(-k/2 \cdot \sum_{i=1}^{n-1}\Delta{(x_i)^2}\right)$$

I didn't get how the density of $x$ was represented in the form of increments. I mean how we can use increments to show the density how are they related or can be used in the density of $x$.

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What are $x$ and $\mathbf{x}$ and how are they related to $\Delta x_i$? It's natural to suppose that $x=\Delta x_1 + \Delta x_2 + \cdots + \Delta x_{n-1}$, but would $\mathbf{x}$ then be the vector $(\Delta x_1, \ldots, \Delta x_{n-1})$? These distinctions are important; they can explain what the derived density is trying to show and how to justify it. –  whuber Jul 5 '12 at 2:20
I have updated what x is. I just don't understand how density of x is given by the increments where xvector=(x1,x2...xn)'. I have also updated what x is. Actually, I got the above derivation from a book called Gaussian Markov Random Field Theory and Applications by Rue and Held. As you said for the above derivation to justify xvector should be equal to (deltax{1},...deltax{n-1}) but I don't think it is the case since xvector=(x1,...xn) –  user34790 Jul 5 '12 at 10:35

Let's take this in short steps.

"$\Delta{x_i} \sim N(0,k^{-1})$" means that (up to a constant of proportionality which we can worry about at the end) the distribution function (PDF) of $\Delta{x_i}$ equals $k^{1/2}\exp{(-(k/2)\Delta{x_i}^2)}$. This is (one) definition of what it means to have a normal distribution with mean $0$ and variance $1/k$.

"Independent" implies the joint ($n-1$-variate) distribution function of all the $\Delta{x_i} = x_{i+1}-x_{i}$ is the product of the individual distribution functions. Exploit a basic property of the exponential (products of its values correspond to sums of its arguments) to write this joint density as

$$k^{1/2}\exp{(-(k/2)\Delta{x_1}^2)}\cdots k^{1/2}\exp{(-(k/2)\Delta{x_{n-1}}^2)}=k^{(n-1)/2}\exp{\left(-k/2(\Delta{x_1}^2+\cdots+\Delta{x_{n-1}}^2)\right)},$$

once more ignoring the constant.

Finally, because each $x_i$ is obtained by starting with $x_1$ and adding all previous increments,

$$x_i = x_1 + \Delta{x_1} + \Delta{x_2} + \cdots + \Delta{x_{i-1}},$$

we can rewrite the preceding sum of squares $\Delta{x_1}^2+\cdots+\Delta{x_{n-1}}^2$ in terms of the $x_i$ to get $x_1^2+2(x_2^2 + \cdots + x_{n-1}^2)+x_n^2 - 2(x_1x_2 + x_2x_3 + \cdots + x_{n-1}x_n)$. Plug this into the preceding formula to obtain the joint distribution of the $x_i$.

One usually doesn't care much about the details of this resulting (somewhat messy) expression; its form is what matters. Apart from constant multipliers, which will be determined by the integrate-to-unity criterion satisfied by any probability distribution, it is the exponential of a quadratic form $Q$. This means the $x_i$ have a multivariate normal distribution. You can read off their variances and covariances by inverting $Q$.

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That, plus a trivial Jacobian. –  Did Jul 18 '12 at 18:19
Since each $Δx_i \sim N(0,1/k)$ (assuming $1/k$ is the standard deviation) then the walk travels $x$ equal to the sum of the $n-1$ independent increments and has a normal distribution $N(0,\sqrt{(n-1)}/k)$ because the variance is the sum of the variances of the individual variances and there are $n-1$ each with variance $1/k^2$. Now $x=∑Δx_i$. So the formula should represent that density. It is not quite written correctly the way you have it. It should be $$\frac{1}{\sqrt{2π(n-1)}/k} \exp \left(-\frac{(∑Δx_i -0)^2}{2(n-1)/k^2}\right)$$