So I have 20 different 'weeks' that are being considered as part of a weight loss exercise. The weight lost each week is distributed normally around mean $\mu =0.8kg$ and std. deviation $\sigma^2= 0.1kg$.
The variance of the total will be the sum of the 20 variances, plus $2\cdot{{20}\choose {2}} \cdot \mathrm{Cov}$. With the correlation between weeks given as $-0.1$, the sum I have attempted as: $(20\cdot0.1^2) + (2\cdot{{20}\choose{2}})\cdot(-0.1\cdot\sqrt{0.1^2 \cdot 0.1^2})$ results in a negative variance of $-0.38$. Could someone please assist me as to where I have gone wrong?
Thanks very much
If the correlation between each pair of successive items amongst $n$ items with common variance $v$ is $c$ and if all the other correlations are zero then the variance $v_n$ of the sum of these $n$ items is $$v_n=nv+2(n-1)cv=(n+2c(n-1))\,v.$$ This is because the square of a sum $\sum\limits_{i=1}^nx_i$ of $n$ terms has $n$ terms $x_i^2$ and $2(n-1)$ terms $x_ix_j$ with $|i-j|=1$. Thus, the expectation of the sum of $n$ random variables in the present situation yields $n$ times their common variance plus $2(n-1)$ times the common covariance $cv$, all the other $(n-2)(n-1)$ covariances being zero.
In the present case, $n=20$, $c=-0.1$ and $v=0.01$ hence the variance of the sum is $$v_{20}=16.2\times0.01.$$ Note in addition that, by the computation above, admissible values of the correlation $c$ are such that $$c\geqslant-\frac{n}{2(n-1)}$$