# Converting a recurrence relationship into a summation

I can find the $i th$ term of a recurrence relationship $Y$ like this:

$$$$Y_i=\sigma_i+\beta_iY_{i-1}$$$$ where $\sigma$ and $\beta$ are normally distributed random variables. I'd like to convert this into summation over $i=1..n$ something like this:

$$$$Y=\sigma_i+\displaystyle \sum_{i=2}^{n-1}\beta_iY_{i-1}$$$$

I know this summation is wrong, any ideas?

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## 1 Answer

Start without the $\Sigma$ in there: $$Y_i=\sigma_i+\beta_i(\sigma_{i-1}+\beta_{i-1}(\sigma_{i-2}+...))= \sigma_i+\beta_i\sigma_{i-1}+\beta_i\beta_{i-1}\sigma_{i-2}= \sum_{j=0}^i \left(\prod_{k=j+1}^i \beta_k\sigma_j\right)$$

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Thank you. It looks beautifully simple, though I'm a little confused by the indices j and k. These are dummy indices? –  HCAI Oct 16 '12 at 23:11
Yeah, they're meaningless. It's like $Y_i$ is a sum over the first $i+1$ $\sigma$s, which are indexed by $j$, and each $\sigma_j$ has a coefficient that's a product of some $\beta$s, which are indexed by $k$-but $k$ starts over for each $\sigma_j$. Try unwrapping the expression for $Y_1, Y_2$ to see better how it works. –  Kevin Carlson Oct 17 '12 at 1:00