# MSW when there is only one observation

The formula for MSW is $$\hat{s}^2_j=\frac{1}{T-1}\sum_{t=1}^{T}w_{jt}(X_{jt}-\bar{X}_j)^2$$ $$\hat{s}^2=\frac{\sum_{j=1}^J\hat{s}^2_j}{J}=MSW$$ where $j$ denotes the risk factor and $t$ denotes the year of observation. There are in total $J$ risk factors and $T$ periods of observation.

Question: When there is only observation such that $T=1$ for $J=3$ risk factors, how is the formula evaluated? A question that I'm doing which involves this situation has solutions which imply that $$\hat{s}^2=\frac{\sum_{j=1}^{J}\sum_{t=1}^{T}w_{jt}X_{jt}}{\sum_{j=1}^{J}\sum_{t=1}^{T}w_{jt}}=\frac{\sum_{j=1}^{J}\sum_{t=1}^{T}w_{jt}X_{jt}}{w_{\Sigma\Sigma}}=\bar{X}$$ I'm unable to see how $\hat{s}^2=\bar{X}$ when $T=1$.

Any help is appreciated.

• You simply cannot estimate a variance with a single observation. – Xi'an Nov 8 '16 at 9:14
• Thank you for your comment. I have missed a detail in my question. $S_{jt}=w_{jt}X_{jt}$ is Poisson distributed. By the mean-variance relationship, it is inferred that the variance equals the mean, where the mean is estimated as above. – none Nov 8 '16 at 10:02