MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $Z_t$ follows a simple discrete random walk $Z_t=Z_{t-1}+e_t$ , where $e_t$ are a bunch of uncorrelated normal variables with arbitrary variance sigma^2, and that there are observations of the series at t=a and t=b, with both observations having normal uncorrelated observational error with variance $O_a$ and $O_b$.

How can I find distribution for intermediate values between a and b?

share|cite|improve this question

if you consider a Gaussian vector $V=(X,Y) \in \mathbb{R}^{d=m+n}$, you know how to find the conditional distribution of $X$ knowing the value of $Y=y$, right ? This is exactly the same thing here.

For example, let us suppose that $a=0, b=N+1$:

  • you have a noisy observation $Y=(y_1, y_2)=(O_a, O_b)$ with know covariance matrix $\Sigma_Y$
  • the data you are looking for, $X=(z_1, \ldots, z_N) \in \mathbb{R}^N$, have a known covariance matrix $\Sigma_X$
  • the covariance matrix $E[X Y^t] = \Sigma_{X,Y}$ is also known.

A quick way to find the conditional distribution of $X$ knowing $Y$ is to write $$X = AU + BV$$ $$Y=CU$$ where $U,V$ are independent standard Gaussian random variable of size $2$ and $N$ respectively, while $A \in M_{N,2}(\mathbb{R})$ and $B \in M_{N,N}(\mathbb{R})$ and $C \in M_{2,2}(\mathbb{R})$. Because

  • $CC^t = \Sigma_Y$ gives you $C=\Sigma_y^{\frac{1}{2}}$,
  • $AC^t = \Sigma_{X,Y}$ then gives you $A=\Sigma_{X,Y}\Sigma_y^{-\frac{1}{2}}$,
  • $AA^t + BB^t = \Sigma_{X}$ then gives you $B=(\Sigma_{X}-\Sigma_{X,Y}\Sigma_y^{-1}\Sigma_{X,Y})^{\frac{1}{2}}$,

the $3$ matrices are easily computable, and $C$ is invertible in the case you are considering. This shows that if you know that $Y=y$, the conditional law of $X | Y=y$ is given by $$X = AC^{-1}y + BV,$$ which is a Gaussian vector with mean $AC^{-1}y = \Sigma_{X,Y}\Sigma_y^{-1}y$ and covariance $BB^t = \Sigma_{X}-\Sigma_{X,Y}\Sigma_y^{-1}\Sigma_{X,Y}$

share|cite|improve this answer
I'm having trouble working through the notation, specifically the covariance matrix of Y (The covariance matrix of the sampling error? Or of the observations?). Could you elaborate the calculation for N=2? – David Shor Apr 11 '10 at 11:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.