# Why does this statistic have an approximate $\chi_{\nu}^{2}$ Distribution?

I am currently studying goodness of fit tests and the $\chi^{2}$ distribution. To calculate the goodness of fit of a theoretical distribution, we compute the quantity $$X^{2}=\sum_{i=1}^{n}\frac{(O_{i}-E_{i})^{2}}{E_{i}}$$ Where $n$ is the number of outcomes, $O_{i}$ is the number of times the $i^{th}$ outcome is observed, and $E_{i}$ is the number of times we expect it to occur (given our distribution).

My book then states $X^{2} \sim \chi^{2}_{\nu}$ approximately, where $\nu$ is an appropriate number of degrees of freedom for the data.

It then defines the $\chi^{2}$ distribution as follows:

If $Z_{i} \sim N(0,1)$ and $X=\sum_{i=1}^{\nu}Z_{i}^2$, then $X\sim \chi^{2}_{\nu}$.

It also briefly proves that $X^{2}$ has an approximately $\chi^{2}$ distribution when $\nu=1$. My question is this:

How do we prove that, for any $\nu$, $X^{2}$ has an approximately $\chi^{2}$ distribution?

Intuitively I feel like this is the start of an induction proof, but I don't know how to show the inductive step. If this is the case, please show me how.

NOTE: I have just noticed that there is an added condition: Each of the $E_{i}$ are greater than or equal to 5.

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The distribution of $X^2$ is not $\chi^2$ (for one thing, $X^2$ can only take a finite number of values). Read your book again. – Did Dec 19 '12 at 12:07
You are right, the book says that this distribution is only approximate. I have changed my question accordingly. – Daniel Littlewood Dec 19 '12 at 12:11