San José State University
Department of Economics
Thayer Watkins
Silicon Valley
& Tornado Alley

The Relative Variance of a Binary Function
of Random Variables Compared to that of
the Corresponding Unary Function

Measured characteristics of males and females have distributions which are at least approximately normal distibutions. Generally the means of these distribution are not significantly different but the variances of the distributions for females is significantly less than those for males. Rodolfo A. Gonzalez has hypothesized that the genes for such characteristics are carried only on the x chromosomes. Females have two X chromosomes, one from each parent, and males only one and that one from his mother. According to Gonzalez' formulation females would have the average of the abilities encoded in their two chromosomes. This reduces the variance compared to that of males. It would make the variance for females equal to one half of the variance for males. The relationship of a female's ability to that of her two parents may be more complicated than a simple average. This is the investigation of the mathematics of the matter.

Let f(u,v) be a function of two random variables, u and v, which have the same distribution. This means that the expected value of u and v are the same, E{u}=E{v}, and their variances, Var(u) and Var(v) are the same. The function f(u,v) is symmetric; i.e., f(v,u)=f(u,v). Let g(u)=f(u,u) be the corresponding unary function.

For small values of the variances of u and v

f(u,v) = f(E{u},E{v}) + fu(u−E{u}) + fv(v−E{v})
and thus
f(u,v)−E{f(u,v)} = [f(E{u},E{v})−E{f(u,v)}]
+ fu(u−E{u}) + fv(v−E{v})

For simplicity let α=[f(E{u},E{v})−E{f(u,v)}].


[f(u,v)−E{f(u,v)}]² = α² + αfu(u−E{u}) + αfv(v−E{v})
+ f²u(u−E{u})² + 2fufv(u−E{u})(v−E{v}) + f²v(v−E{v})²

Since E{u-E{u}}=0 and E{v-E{v}}=0 the taking of the expected values in the above equation yields

Var(f) = α² + f²uVar(u) + 2fufvCov(u,v) + f²vVar(v)

If u and v are independent then Cov(u,v)=0. The above equation then reduces to

Var(f) = α² + f²uVar(u) + f²vVar(v)

Let Var(u) and Var(v) be denoted as σ². Since E{u}=E{v} and f is symmetric fu(E{u}) is equal to fv(E{v}). Let the common value be denoted as φ. As σ² → 0, α → 0, so take α to be zero. The above equation is then

Var(f) = 2φ²σ²

On the other hand if u and v are the same, as in g(u)=f(u,u), then Cov(u,v)=σ². This means that

Var(g) = φσ² + 2φσ² + φσ² = 4φσ²
and therefore
Var(f)/Var(g) = 1/2

The ratio of the standard deviations is then 1/√2, approximately 0.7.

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