I am trying to measure and quantify risk, variance, and standard deviation over a time period $T$. It is broken into two sub-periods $t_1$ and $t_2$. $X_1$ is the time series for $t_1$, and $X_2$ is the time series for $t_2$, where $t_1+t_2=T$. We can assume the $X_1$ and $X_2$ are sub-time series of a longer time series. All I know is that the variances over the two sub time periods are, say $S_1$ for $t_1$ and $S_2$ for $t_2$; we do not know what $X_1$ and $X_2$ are.

I want to find the variance over the entire time period $T$ if $S_1$ is the variance for time period $t_1$, and $S_2$ is the variance for time period $t_2$, and $t_1 + t_2 =T$. My rationale so far:

If $S_1=S_2=S$, then the variances over $t_1$ and $t_2$ are $S_1t_1$ and $S_2t_2$ respectively. Since $S_1=S_2$, it simply becomes $S_1T$ as $t_1+t_2=T$. The standard deviation over the time period is $S\sqrt{T}$(squaring both terms and adding, then take the square root again).

If $S_1$ and $S_2$ are not the same, then is the formula for standard deviation:
$$ s_T=\sqrt{S_1t_1 + S_2t_2 + 2\sqrt{t_1t_2}{\rm Cov}(X_1, X_2)} $$ or do I add like it is a portfolio, with result being:
$$ s_T=\sqrt{\frac{S_1t_1^2}{T^2} + \frac{S_2t_2^2}{T^2} + \frac{2t_1t_2}{T^2}{\rm Cov}(X_1, X_2)} $$ where $s$ is the standard deviation?

I have been told to use square root of time to change time periods for standard deviation. However, if we think of the time series as a portfolio with weightings $t_1/T$ and $t_2/T$, then the weights are not the square root of time with respect to variance but just the straight time weights as in the portfolio risk formula: $$ s_p=\sqrt{\frac{S_1h_1^2}{H^2} + \frac{S_2h_2^2}{H^2} + \frac{2h_1h_2}{H^2}{\rm Cov}(X_1, X_2)} $$ where $\frac{h_1}{H}+\frac{h_2}{H}=1$

I hope this clarifies what I am asking. Can someone please explain to me the proper way to think about this? Do I add variances with time squared or just time?

  • $\begingroup$ I took the liberty of editing your question with $\LaTeX$ to make it easier to read. Please ensure it still says what you want it to. $\endgroup$ – gung Aug 25 '14 at 2:29
  • $\begingroup$ What exactly do you mean by a "variance measurement"? How might it be related to "$\text{Cov}(X_1,X_2)$? $\endgroup$ – whuber Aug 25 '14 at 14:02
  • $\begingroup$ Given $S_1$ and $S_2$ as variances over $t_1$ and $t_2$ I want to find the variance formula for $T=t_1+t_2$. $\endgroup$ – Sean vader Sep 2 '14 at 2:49

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