# Expected Time Until Absorption and Variance of Time Until Absorption for absorbing transition matrix P, but with a Probability Vector u

I didn't see this topic covered in the prompts. I know how to find the fundamental matrix, N, of an absorbing transition matrix, P. This has also already been covered a lot in prior posts. However, I'm having difficulty finding how to treat this when introducing a probability vector, u, for the starting state. Would it simply be following the usual steps to find N, but using a canonical form of uP?

EDIT: If $$u$$ is instead a $$nxn$$ probability matrix, and $$N$$ is the $$nxn$$ fundamental matrix with $$0$$s for all absorbing states, then the Expected Time Until Absorption for each state can be found by $$t^{u} = u \times N \times \textbf{1}$$. Typically (without a probability matrix), the Expected Time Until Absorption for each state is found by $$t = N \times \textbf{1}$$, and the Variance is found by $$(2N - I) \times t - t_{sq}$$, where $$t_{sq}$$ is each element of $$t$$ squared. In an attempt to adjust the variance to account for the probability matrix $$u$$, I tried $$(2u \times N - I) \times t^{u} - t^{u}_{sq}$$. This is evidently not correct since there exists probability matrices $$u$$ such that $$t^{u}_{sq}>(2u \times N - I) \times t^{u}$$, leading to a negative variance. What would be the correct way to calculate variance in time until absorption adjusting for a probability matrix?

• Ordinarily you find the expected absorption time, say $t(x)$, starting from each state $x$. Given an initial probability vector $u(x)$, you can reconstruct the expected absorption time from $t(x)$ by using the formula $\sum_x u(x) t(x)$. Or if you use the usual convention that probability vectors are row vectors and function vectors are column vectors, this is $ut$.
– Ian
Commented Aug 4, 2020 at 18:38
• @Ian Thanks! that does make more sense. Clarifying question: would $u$ and $t$ have row/column length equal to the number of transient states? Commented Aug 4, 2020 at 18:48
• You would usually have them encompass the entire state space, but $t(x)=0$ for all the absorbing states $x$, so only the component of $u$ concentrated on the transient states will actually contribute anything to the product.
– Ian
Commented Aug 4, 2020 at 18:52
• @Ian If you have time, I ended up running into another issue while working on a problem. My first attempt didn't make sense. Please see my edits. Thanks. Commented Aug 5, 2020 at 17:36
• I don't think you can find the variance in absorption time started from each state by only knowing the expected value of that absorption time. Can you give a reference for where you got $(2N-I)(t-t \cdot t)$ (where $\cdot$ is the elementwise product)? Also can you explain how this formula works with, for example, a deterministic process $A \to B \to C$?
– Ian
Commented Aug 5, 2020 at 19:58

Regarding your first question, given the expected time $$t(x)$$ to absorb starting from each state $$x$$ and the initial probability distribution $$u(x)$$, the expected time to absorb with this initial probability distribution is $$\sum_x u(x) t(x)$$. Normally we write distributions as row vectors and functions as column vectors; in this case this can be written as just $$ut$$. In terms of the fundamental matrix, $$t=N \mathbf{1}$$ so this is $$uN\mathbf{1}$$.

Regarding your question about the variance, here is a way to do it from scratch. You have the recursion:

$$E[\tau^2 \mid X_0=x]=\sum_y E[(\tau+1)^2 \mid X_0=y] P[X_1=y \mid X_0=x]$$

which expands to

$$E[\tau^2 \mid X_0=x]=\sum_y E[\tau^2 \mid X_0=y] P[X_1=y \mid X_0=x] + 2 \sum_y E[\tau \mid X_0=y] P[X_1=y \mid X_0=y] + 1.$$

Thus the conditional variance can be obtained as

$$V[\tau^2 \mid X_0=x]=\sum_y E[\tau^2 \mid X_0=y] P[X_1=y \mid X_0=x] + \sum_y E[\tau \mid X_0=y] P[X_1=y \mid X_0=y]$$

for each non-absorbing state $$x$$. This follows by looking at the recursion for the expectation:

$$E[\tau \mid X_0=x]=1 + \sum_y E[\tau \mid X_0=y] P[X_1=y \mid X_0=y]$$

and subtracting on both sides. In matrix notation you could write the equation for the variance as

$$v(x)=(P(v+t))(x)$$

for each non-absorbing state $$x$$ and $$v(x)=0$$ for the absorbing states. You can then again multiply out $$uv$$ if you have an initial probability distribution $$u$$.

Regarding your question about computing the variance using the fundamental matrix, I looked into this, and it can be done. But you need to be careful about the grouping. The variance started from each state is given by the column vector

$$(2N-I)t-(t \circ t)$$

where $$\circ$$ is the Hadamard product. You can multiply that whole column vector by $$u$$ on the left to get the variance of the absorption time for an initial probability distribution $$u$$. You do not replace $$t$$ by $$ut$$ everywhere in the identity above.

• Functionally this is working just fine! Out of curiosity, do you have a citation for this? Only reason I had to ask on here is because I can't seem to find any books/papers addressing this exact problem. Once again, thanks for the help! Commented Aug 7, 2020 at 13:25
• @R.Story For the first thing I just picked this up in coursework; it comes down to the total expectation formula in the end. Though various calculations like this are under the umbrella of "renewal theory".
– Ian
Commented Aug 7, 2020 at 14:08