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I'm interested in the proportion of time that a sensor is in a particular state. The sensor tells me the amount of time that it's in each state, which I will denote by $X = \{ X_1, X_2, X_3\}$. I also set the interval time over which the sensor can be in any one of those states, which is a known duration that I will denote by $T$. $$ T = \sum_{i = 1}^3 X_i $$ I'm really interested in the proportion of the known interval time, $T$, that the sensor is in each state, $X$. $$ p_i = \dfrac{X_i}{T} $$ I am measuring $X$ and know $T$. The sensor also outputs the $\text{sd}(X_1)$ and the $\text{sd}(X_2)$, but not the $\text{sd}(X_3)$. I need to figure out how to calculate the $\text{sd}(X_3)$??? Here $\text{sd}(\cdot)$ refers to the standard deviation (which in this case is in units of time). This is kind of like a Dirichlet distribution (thx to the clarification from @jbowman). Any advice on how to treat this situation? I basically need to calculate $X_1 + X_3$ (which is simple b/c I know both of these values) and the $\mathrm{Var}\left(X_1 + X_3 \right) \stackrel{?}{=} \mathrm{Var}(X_1) + \mathrm{Var}(X_3)$ (which is a little more difficult b/c I know one of the values, I just need to calculate the other given the constraints and make an assumption that the sum of the variances is the variances of the sum).

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    $\begingroup$ This is not a multinomial distribution, which is discrete and on the integers. It most resembles a Dirichlet distribution. I don't see how, if you have the measurements of $X_1$ etc., there is a standard deviation involved; if $X_1 = 4, X_2 = 6, T = 15$, where is there any variability in $X_1$? $\endgroup$
    – jbowman
    Commented Feb 5, 2018 at 15:51
  • $\begingroup$ @jbowman - the data from the sensor would be $X_1 = 4 \pm 1$ that might read 4 seconds plus / minus 1 second. So perhaps I should of framed the $X$s as averages. All the $X$s must sum to $T$, but there is some wiggle room there. Does that clarify things? $\endgroup$
    – user13317
    Commented Feb 5, 2018 at 17:14
  • $\begingroup$ Yes, mostly. Is $T$ known with 100% accuracy, or does it have a std. deviation too? $\endgroup$
    – jbowman
    Commented Feb 5, 2018 at 17:16
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    $\begingroup$ You don't have enough information. The reason is that since $X_3=T-(X_1+X_2)$, the algebraic properties of variances imply $$\operatorname{sd}(X_3)^2=\operatorname{Var}(X_3)=\operatorname{Var}(T-X_1-X_2)=\operatorname{Var}(X_1+X_2)\\=\operatorname{Var}(X_1)+\operatorname{Var}(X_2)+\operatorname{Cov}(X_1,X_2)=\operatorname{sd}(X_1)^2+\operatorname{sd}(X_2)^2+\operatorname{Cov}(X_1,X_2).$$ You need information about this covariance (or equivalently about the correlation between $X_1$ and $X_2$). $\endgroup$
    – whuber ♦
    Commented Feb 7, 2018 at 20:45
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    $\begingroup$ @whuber Thank you! At one point I tried to use the equation for the variance of a Dirichlet distribution to work out an approximation, but came to the same conclusion that there were too many unknowns. Your approach is way more eloquent, and much appreciated! $\endgroup$
    – user13317
    Commented Feb 8, 2018 at 16:13

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