# X~Unif(0, 1) ; X1 + X2 + … X6 = 1 ; Y = sum(X1…X6) ; VAR(Y) =?

Let $$X_i$$ ~ Unif(0, 1) s.t.

$$X_1 + X_2 + ... + X_6 = 1$$

Let $$Y = X_1 + … X_6$$

What is $$Var(Y)$$?

(Also the case when it's $$X_n$$)

Purpose for the curious:

I'm trying to rank confidence for softmax predictions using variance.

6 classes,

1 | .01, .01, .01, .01, .01, .95 ; var=.1227, max=.95
2 | .01, .01, .01, .01, .50, .46 ; var=.0492, max=.50
3 | .01, .01, .01, .33, .33, .30 ; var=.0241, max=.33
4 | .01, .01, .25, .25, .25, .23 ; var=.0123, max=.25
5 | .01, .20, .20, .20, .20, .19 ; var=.0049, max=.20
6 | .16, .16, .16, .16, .16, .20 ; var=2.2e-4, max=.20
… etc

If we rank them by max or var, will both list be same?

Hypothesis: Similar but no.

In this case,
- Var would rank obs#5 higher than #6; bc var says there’s no way the pred can class 1 in #5
- Max, on the other hand would say there’s no difference in information between #5 and #6

For cutoff idk if the variance of Unif(0, 1) would be a good cutoff or if I need to solve the problem I mentioned above for insight.

• As your question is written, $\operatorname{Var}(Y)=0$. If you have a random vector $X:=(X_1,\dots,X_n)$ constrained to sum to $1$ then the variance of the sum is zero since it's constant – jld May 8 '20 at 17:29
• Wait I'm dumb... what am I trying to ask then? Am I trying to find Var of X? I think I've gone full circle. – Linsu Han May 8 '20 at 17:31

jld has already shown in his comment that $$Var(Y)=0$$ because $$Y=0$$ and a constant must have variance zero.

It should also be noted that the question is only possible when $$n=1$$.

From the question, generalized to n:

Let $$X_i$$ ~ Unif(0, 1) s.t.

$$X_1 + ... + X_n = 1$$

That is, each $$X_i$$ is an uniform variable but the sum of all $$X_i$$ is 1. Since the expectation of sum is sum of expectations and all $$X_i$$ have the same distribution:

$$E(X_1 + ... + X_n) = n·E(X_i)$$

and

$$E(X_1 + ... + X_n) = E(1) = 1$$

and

$$E(X_i)=0.5$$ because $$X_i$$ ~ Unif(0, 1)

$$n=\frac{1}{0.5}=2$$

Furthermore, we can compute $$Cov(X_1,X_2)$$ and $$Cor(X_1,X_2)$$

$$0=Var(1)=Var(X_1+X_2)=var(X_1)+Var(X_2)+2\cdot Cov(X_1,X_2)=$$

$$=\frac{1}{12}+\frac{1}{12}+2\cdot Cov(X_1,X_2)$$

$$Cov(X_1,X_2)=-\frac{1}{12}$$

$$Cor(X_1,X_2)=\frac{Cov(X_1,X_2)}{\sqrt{(Var(X_1)\cdot Var(X_2))}} =\frac{-\frac{1}{12}}{\frac{1}{12}}=-1$$

In fact, that matches the only solution of two random variables with the same distribution and constant sum $$X_2=1-X_1$$

And returning to the question (modified in comments) after that digression: $$Var(X_i)=\frac{1}{12}$$ just as with any uniform variable.

And interestingly, the sample variances in the original question are different from 1/12 because those samples don't add to 1 or are uniform.