Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We have a sequence $\lbrace b \rbrace$ which is defined in a relation $b_i = b_{i-1} * 0.9 + t_i * 0.1$. If we have observations of the full sequence of $\lbrace b \rbrace$, it would be quite simple to calculate the values of $\lbrace t \rbrace$. However, the case is that we only have around $50\%$ to $80\%$ observations in $\lbrace b \rbrace$ and we still want to calculate the values of $\lbrace t \rbrace$ as accurately as possible. Are there any well-known mathematical methods to solve this problem? Thanks!

share|improve this question
What do you mean by "now"? What did we have before? –  Yuval Filmus Mar 1 '12 at 21:15
@YuvalFilmus It's more like a misconception...I should edit it... –  derekhh Mar 1 '12 at 21:16

1 Answer 1

up vote 3 down vote accepted

The simplest would be to decide that if you know $b_i$ and $b_j$ for $i<j$ but none of the intermediate indices, then $t_{i+1}$ through $t_j$ will be estimated as the same value. It is then simple to telescope the equations and find that common value given $b_i$ and $b_j$.

Whether this is a reasonable model depends on how your data were produced, of course.

In some applications, it may be better to require that $t_i$ through $t_j$ is an arithmetic (or even geometric) sequence. Assuming that you already know $t_i$, you can again telescope the definitions of $b_{i+1}$ to $b_j$ and find how much of a per-step increase in $t$ (or multiplier on $t$) you need for $b_j$ to come out right.

Beware that the latter method carries a risk that if there are long runs where you know only isolated values of $b_i$, the entire sequence of $t$ estimates will be sensitive to the first known $b$, because you're using already-estimated $t_i$s to produce new ones. Thus, you could end up with wildly oscillating $t$s even for $b$s that don't vary much. In contrast, the "equal $t$s" approach don't depend on anything but the two neighboring $b$s and is therefore more stable.

share|improve this answer

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.