# partitioning variances, higher order products and multivariate skewness

If A,B and C all contribute to Z as:

$$Z = A*B*C$$

we can get the contributions of ABC to the variance in Z as:

$$Var(Z) = Var(ABC)$$

getting independent contributions of A, B and C and covariances requires:

$$Var(Z) = \bar{B}^2\bar{C}^2Var(A) + \bar{A}^2\bar{C}^2Var(B) + \bar{A}^2\bar{B}^2Var(C) + 2\bar{A}\bar{B}\bar{C}^2Cov(A,B) + + 2\bar{A}\bar{B}^2\bar{C}Cov(A,C) + 2\bar{A}^2\bar{B}\bar{C}Cov(B,C) + Q$$

where $Q$ is a remainder term calculated by subtracting all everything other parameter in the third equation from $Var(Z)$.

In some cases the remainder term can be large, and I have read that the remainder can reprint multivariate skewness and high order terms such as products of variances and covariances.

I would like to fully grok the idea of remainder terms being built upo by higher order terms and skewness.

What are these higher order terms and what do the represent and why does multivariate skewness explain some of the variation in Z?

(could this be a maths overflow question?)

• I can't really tell which "higher order terms" you refer to, but it strikes me that when $A$ and $B$ are iid standard Normal and $C$ is independent of them both (and therefore could be chosen to make $Z$ have enormous variance), then your second formula reduces to $\text{Var}(Z)=Q$, which seems of little value. – whuber Jan 6 '15 at 17:14
• Your first formula is wrong: it is the one for $A+B+C$! – Xi'an Jan 6 '15 at 17:33
• so you are saying $Var(ABC)$ cannot be expanded into $\bar{B}^2\bar{C}^2Var(A) + \bar{A}^2\bar{C}^2Var(B) + \bar{A}^2\bar{B}^2Var(C) + 2\bar{A}\bar{B}\bar{C}^2Cov(A,B) + 2\bar{A}\bar{B}^2\bar{C}Cov(A,C) + 2\bar{A}^2\bar{B}\bar{C}Cov(B,C)$ ? – user1317221_G Jan 7 '15 at 14:14
• George W. Bohrnstedt and Arthur S. Goldberger Journal of the American Statistical Association Vol. 64, No. 328 (Dec., 1969), pp. 1439-1442 would seem to suggest otherwise – user1317221_G Jan 7 '15 at 14:31

As Xi'an has noted (see comments to question), the OP's formula for $Var(Z)$, when $Z = A*B*C$, is incorrect.
According to the developmental version of mathStatica, the variance of $Z = A*B*C$ ... expressed in terms of the moments of $A$, $B$ and $C$ ... is:
where $$\mu _{r,s,t} = E\left[ \left(A -E\left[A\right]\right)^r \left(B-E\left[B\right]\right)^s \left(C -E\left[C\right]\right)^t \right]$$
If $A$, $B$ and $C$ are independent, this result simplifies to:
$$Var(A B C) = E\left[A^2\right] E\left[B^2\right] E\left[C^2\right] -(E[A] * E[B] * E[C])^2$$
• so you are saying $Var(ABC)$ cannot be expanded into $\bar{B}^2\bar{C}^2Var(A) + \bar{A}^2\bar{C}^2Var(B) + \bar{A}^2\bar{B}^2Var(C) + 2\bar{A}\bar{B}\bar{C}^2Cov(A,B) + 2\bar{A}\bar{B}^2\bar{C}Cov(A,C) + 2\bar{A}^2\bar{B}\bar{C}Cov(B,C)$ ? – user1317221_G Jan 7 '15 at 14:05