# Mean of the square of a sum of exponential terms

1. Oct 20, 2015

### grepecs

1. The problem statement, all variables and given/known data

Calculate $\widehat{Y^{2}}$

(i.e., the mean of the square of $Y$.

2. Relevant equations

$$Y=\sum_{k=0}^{N-1}y_{k}$$

where

$$y_{k}=e^{-\gamma t}e^{\gamma \tau k}G_{k}$$

and

$$t=N\tau$$

The quantities $y_{k}$ (or $G_{k}$) are statistically independent.

3. The attempt at a solution

$$\widehat{Y^{2}}=\widehat{G^{2}}e^{-2\gamma t}\sum_{k=0}^{N-1}e^{2\gamma \tau k}=\widehat{G^{2}}e^{-2\gamma t}(\ \frac{1-e^{2\gamma t}}{1-e^{2\gamma \tau}} )$$

However, the correct answer should be

$$\widehat{Y^{2}}=\widehat{G^{2}}\frac{1}{2\gamma\tau} (\ 1-e^{-2\gamma \tau} )$$

so it seems like I'm doing something wrong, or is it possible to somehow simplify my answer in order to get the correct one?

2. Oct 20, 2015

### Ray Vickson

Unless the quantities $G_k$ have mean zero ($E G_k = 0$) and are identically distributed for different $k$ (or, at least, all have the same second moments), the expression you give is incorrect; and in that case the one given in the book (or wherever) is also incorrect. As far as I can see, the only way you could obtain something like the second expression would be to have non-zero values of $E G_j G_k$ of the form $E G_j G_k = E(G^2) c_{jk}$, where $E(G^2) \equiv E(G_1^2) = E(G_2^2) = \cdots = E(G_{N-1}^2)$. This is because we have $Y = \sum_{k=1}^{N-1} w_k G_k$ (with $w_k = e^{-\gamma t} e^{\gamma \tau k}$), so that
$$Y^2 = \sum_{k=1}^{N-1} w_k^2 G_k^2 + 2 \sum_{1 \leq j < k \leq N-1} w_j w_k G_j G_k,$$
so you would need an appropriate value for the expectation of the second summation above, and that would depend on the form of the coefficients $c_{jk}$.