I have several nonlinear curves of the transmission spectrum of a semiconductor. I got a fit to each of the curve. To get the chi squared for each my teacher gave me the formula $$\chi^2=\sum_{i=1}^{N}\frac{(T_{exp}-T_{fit})^2}{N\sigma^2_{exp}} $$ where $$\sigma_{exp}=\sum_{i=1}^{N}\frac{\sigma_{i}}{N}.$$ However, reading some statisics book I haven't seen this formula anywhere and have only seen $$\chi^2=\sum_{i=1}^{N}\frac{(T_{exp}-T_{fit})^2}{\sigma^2_{i}}.$$ Can someone explain the top, or is the formula just wrong?
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$\begingroup$ Pull the $1/N$ out of the summation. Then you see it's just a scaling-by-N factor. $\endgroup$– Mike DunlaveyOct 8, 2013 at 12:02
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$\begingroup$ That wouldn't be the same you would get $$\chi^2=\sum_{i=1}^{N}\frac{N(T_{exp}-T_{fit})^2}{\sum_{i=1}^{N}\sigma_{i}} $$. $\endgroup$– CrazyDolphinOct 8, 2013 at 12:46
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$\begingroup$ Edit: Denominator should be squared. $\endgroup$– CrazyDolphinOct 8, 2013 at 13:02
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$\begingroup$ $$\chi^2=\frac{1}{N}\sum_{i=1}^{N}\frac{(T_{exp}-T_{fit})^2}{\sigma^2_{exp}}$$ $\endgroup$– Mike DunlaveyOct 8, 2013 at 13:22
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$\begingroup$ What exactly do you mean by $\sigma_i$ in the expression for $\sigma_{exp}$? Regardless, this formula is unlikely to be a good estimate of the standard deviation of the residuals. $\endgroup$– whuber ♦Oct 8, 2013 at 17:20
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1 Answer
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Let's unpack the formula. Ignoring the controversial factor of $1/N$, it is a sum of expressions of the form
$$\frac{(T_{exp}-T_{fit})^2}{\sigma^2_{exp}}.$$
These are squares of ratios. The numerators, $T_{exp}-T_{fit},$ are the residuals: they (additively) compare the observed values (presumably $T_{exp}$) to the fitted values $T_{fit}$. This makes sense when the hypothesized explanation for any differences between the fit and the observations is some random, uncorrelated additive variation.
The denominators, $\sigma_{exp},$ are estimates of the expected amount of variation (based on the fit). In some circumstances they vary from point to point; in other situations they are taken as constant. No matter: the point is that the ratio
$$\frac{T_{exp}-T_{fit}}{\sigma_{exp}}$$
re-expresses each residual as a multiple of the expected amount of variation. (Notice that it is unitless.) Intuitively, then, these ratios ought to be around $\pm 1$ in size, although we would hope many of them would be closer to $0$. If we make a number of statistical assumptions, have sufficient amounts of data, perform least-squares or maximum likelihood fitting, and invoke the Central Limit Theorem, we may justify considering these ratios as being approximately, sort of, perhaps, Normally distributed.
The sum, then--without the offending factor of $1/N$ that would turn it into a mere average--looks like the sum of squares of uncorrelated standard Normal variables. This is the definition of $\chi^2$. That is, when all these assumptions hold, we may interpret the sum
$$\sum_i\frac{(T_{exp}(i)-T_{fit}(i))^2}{\sigma^2_{exp}(i)} = \sum_i\left(\frac{T_{exp}(i)-T_{fit}(i)}{\sigma_{exp}(i)}\right)^2$$
(where the index $i$ is used to enumerate the observations) as if it were a single random draw from a $\chi^2$ distribution. Actually that's not quite right, because these residuals are correlated. They are connected by virtue of the common underlying fit, which typically relies on estimates of a small number of parameters, say $p$ of them. Because of this, the sum--qua random variable--actually behaves more like a sum of $N-p$ squared standard Normal variates. This value, $N-p$, is the number of "degrees of freedom."
Were we to divide the sum by $N$, we would obtain a scaled $\chi^2$ distribution. There's nothing wrong about this, but it isn't too helpful, either: tables and software compute values of $\chi^2$ distributions, not values divided by $N$. After all, $N$ isn't even directly relevant: only $N-p$ matters in calculating $\chi^2$. Leaving out the $1/N$ factor therefore makes the calculation depend only on one number, the degrees of freedom, rather than a pair of numbers. That--the second version in the question--is the one to use.
