Puzzle:
Let X and Y be two independent "normal" random deviates with mean
and standard deviation
μ_{X} and σ_{X},
and
μ_{Y} and σ_{Y},
respectively.
These formulas are extremely useful for working with statistics (meaning "numbers equipped with 1σ error bars"). Surprisingly, the multiplication, division, and f formulas do not seem to be given by statistics books.
By Warren D. Smith Sept 2009
A and B are well known. The formulas for the mean arise from linearity of expectations. The formulas for the variance then also arise from linearity of expectations combined with independence. The underlying reason X+Y is exactly normally distributed (ditto X-Y) is that (i) the Fourier transform of a Gaussian is a Gaussian and (ii) the product of two Gaussians is a Gaussian.
For part C, the formula for the mean arises from
[with integration over the whole of the zq plane, and with F(z)=(2π)^{-1/2}exp(-z^{2}/2)] whereupon the formula for standard deviation may be verified by computing
For part D regard μ_{Y}/Y as approximately normal then apply part C's multiplication formula.
The fact that 1/(1+δ)=1-δ+δ^{2}-δ^{3}+... is used to deal with small relative deviations δ off mean (and we pretend the "lower tail of Y is cut off," i.e. that the chance of a deviation of the same order as Y's mean, or larger, is neglectible – otherwise the standard deviation of X/Y would, in fact, be infinite).
This technique plus the fact that ∫_{-∞<s<∞}F(s)s^{2n}ds=1·3·5···(2n-1) if n≥1 allows us to see that the mean of 1/Z, where Z is a normal deviate with mean=1 and standard deviation σ<<1, is given asymptotically when δ→0+ by any truncation of
Meanwhile, using the fact that
we find that the variance of 1/Z is given asymptotically by any truncation of
E. I recommend (following a common recommendation) summing these series up to the term of least absolute value, then stopping (or stopping earlier, if the term is small enough that you're satisfied). The first neglected |term| will then presumably be about the same as the additive error. For each of the δ-power series above we'd keep about δ^{-2}/2 terms. This (as a result of "Stirling's formula," essentially) would yield least |term| roughly (δ/r)^{c+1/δ²} where r=√e=1.6487... and c=O(1) is a constant you can choose to optimize performance of the estimate. (Different c's for different series.)
F. Same technique as in part D but employing the Taylor series of f(z) about z=μ. The odd-power terms in the series integrate to zero due to odd symmetry. When you multiply an even term f^{(2n)}(μ)/(2n)! times the normal-moment-factor 1·3·5···(2n-1)·σ^{2n} you get 2^{-n}/n! f^{(2n)}(μ) σ^{2n}. That explains the first series. The second series is the same series but based on f(z)^{2} not f(z). If f(z)=exp(z) or cos(z) then both series converge for all σ; but as we saw in part D, if f(z)=1/z then we get divergence for all σ≠0. [And the fact that mean^{2}+variance=meansquare for a probability distribution, is just Pythagoras's theorem, integrated.]
G. 2.0925±0.540. [Using δ=0.2 and δ^{2}=0.04 and truncating as in part E.]