# Systematic Component of Variation

The appendix of the paper of McPherson et al (1982) (see screenshot below) contains a derivation of the Systematic Component of Variation (SCV). I understand the derivation with exception of the first step. Here are the premises:

$O_i$: observed cases in region i
$E_i$: expected cases in region i
$\lambda_i$: multiplicative factor associated with region i ($O_i=\lambda_i*E_i$)

Now the following assumptions have been made:

$O_i$ is approximately Poisson distributed with mean $\lambda_iE_i$
$\lambda_i$ is considered as a random variable with expected value $1$ and variance $\sigma^2$.

From these the following formula is concluded:

var($O_i$) = $E_i^2\sigma^2$ + $E_i$

I tried to find out how to get the formula by the given premises and assumptions and didn't succeed. Any idea?

Screenshot of McPherson's derivation:

• I'm a bit surprised that this simple question has languished 8 months without an answer. Dec 16, 2016 at 16:09
• It looks more like 33 months to me, wasn't it asked in April 2014 or am I misreading it? I only came across it by chance as I'm getting my (decidedly non-mathematical) brain around the technique for work. Dec 16, 2016 at 16:19
• @kjetil and Tumbledown: I put the question in October 2016. There is something wrong in the time setting of SO. Dec 17, 2016 at 15:10
• @Tumbledown "Apr 14" means the 14th day of April in the current year. If an earlier year than the current one is intended, it is shown using a format like: Jan 26 '12 (26th day of January, 2012) Dec 17, 2016 at 22:14
• I was convinced I aksed the question in October 2016. Maybe I was wrong. @Glen_b Why not be consequently using mmm dd yyyy? (My favorite would be yyyy-mm-dd or dd.mm.yyyy). Dec 18, 2016 at 14:46

This is simply the law of total variance: https://en.wikipedia.org/wiki/Law_of_total_variance

You have given:

$E_i$ is a known constant

$\DeclareMathOperator{\E}{\mathbb{E}}\E \lambda_i = 1$

$\DeclareMathOperator{\V}{\mathbb{V}}\V \lambda_i=\sigma^2$

$\E(O_i | \lambda_i)=\lambda_i E_i$

$\V(O_i | \lambda_i) = \lambda_i E_i$

Using this with the law of total variance we get: $$\V (O_i) = \E \V (O_i | \lambda_i) + \V \E (O_i | \lambda_i) \\ = \E (\lambda_i E_i) + \V (\lambda_i E_i) \\ = E_i \E(\lambda_i) + E_i^2 \V (\lambda_i) \\ = E_i + E_i^2 \sigma^2 \\ = E_i (1+\sigma^2 E_i)$$

• I'm very happy that this question got answered. Thank you very much. Indeed, the mathematical arrangement is simple. I found the statistical assumptions difficult. The important point for me in your answer was: $E_i$ is a known constant. But isn't $\mathbb E(O_i) = \lambda_i E_i$? Dec 17, 2016 at 15:26
• The question in comment : NO, $\E (O_i | \lambda_i)=\lambda_i E_i$, but $\E O_i = E_i$, since $\E \lambda_i=1$. Dec 17, 2016 at 15:30
• You saved my day. Dec 19, 2016 at 7:41
• @kjetilbhalvorsen They should have been really more verbose. The equation after "so that" is also strange given that the left hand side is a (non-degenerate) random variable, while the right hand side is a deterministic value, and also the next line is a simple rearrangement, yet $\sigma$ is miraculously replaced with $\widehat{\sigma}$. May 26, 2020 at 13:53