Calling all statisticians.
 

Consider the area of a square of width w +/- dw and height h +/- dh. We wish to calculate A = hw +/- dA. How do we calculate dA? Well, we can expect the actual value of A to lie in the range:

(w-dw)(h-dh) < A < (w+dw)(h+dh)

We could define a dA- to be hw - LHS and dA+ to be RHS - hw. Average the two and call it dA and you find:

dA = hdw + wdh.

Or,

dA/A = dw/w + dh/h

However, this is NOT the result of the magic formula-from-a-can for the propagation of uncertainty as annointed by some voodoo panel of statisticians somewhere:

(dA/A)^2 = (dw/w)^2 + (dh/h)^2

Now, I am fully confident that my derivation is somehow incorrect. I would just like to understand the logic behind the correct formula, since the derivation seems so simple.

Any help? The wiki article of propagation of uncertainty just gives the standard formulae, without elaborating on where they came from.

Re: Calling all statisticians.
 

Here is why DA=wdh+hdw is wrong

Label the rectangle A with vertices (0,0) (w,0) ,(0,h) (w,h)

Now Label the larger rectangle which vertices (0,0) (w+dw,0) (0,h+dh) (w+dw,h+dh)

Label the smaller rectangle with vertices (0,0) (w-dw,0) (0,h-dh) (w-dw), (h-dw)

Now we use an incorrect fact that
2DA = the larger rect - the smaller rectangle
2DA= (w+dw)(h+dh)- (w-dw)(h-dh)
2DA= wh +wdh +hdw +dwdh -[wh-wdh -hdw +dwdh]
2DA = 2wdh +2hdw
divide both sides by 2
DA= wdh +hdw

Re: Calling all statisticians.
 

Let W be normal with mean w and standard deviation dw. Let H be normal with mean h and standard deviation dh. Suppose W and H are independent. Let X=WH. Define A = EX and let dA be the standard deviation of X.

By independence, A=wh. Also,

(dA)^2 = E(X^2) - (EX)^2
= E(W^2)E(H^2) - w^2 h^2.

Note that (dw)^2=Var(W)=E(W^2)-w^2. Hence, E(W^2)=w^2+(dw)^2. We therefore have

(dA)^2 = (w^2+(dw)^2)(h^2+(dh)^2) - w^2 h^2.

The (dw)^2(dh)^2 term is asymptotically negligible, so this yields the correct formula.

Edited to add:
Here's another way to look at it. The term dw is the standard deviation of the error in the measurement of the width. Let E_w be the actual error. Your derivation simply reproduces the standard linear approximation

E_A = wE_h + hE_w.

This is valid. But dA is the standard deviation of the left-hand side. By independence, the variance of the sum on the right is the sum of the variances. But the same is not true for standard deviations.

Re: Calling all statisticians.
 

jason1990,

Thanks for the thoughtful analysis.

Re: Calling all statisticians.
 

[ QUOTE ]



(dA)^2 = E(X^2) - (EX)^2


[/ QUOTE ]

Both squares and cubes are in the standard character set.
x² and x³

Re: Calling all statisticians.
 

I should mention, in passing, that the "|hdw| + |wdh|" formulation is not completely unknown. It measures maximum possible error, if everything is off in the worst possible direction at once. You'll see this in computer science, not in statistics: that kind of analysis has to be done to guarantee that a program using floating-point operations to do large-integer arithmetic always gives the correct answer, for example.

The usual squared-uncertainties formulation, as already noted, is based on the assumption that each error is independent.