Square of Binomial is Chi-Square? my statistics professor said in passing that if you square the binomial test statistic, you can derive the chi-square statistic. I tried working out the proof, but I can't seem to reduce the chi-square statistic to one fraction. Does anyone have a proof or a link to the proof? I can't seem to find one.
Thanks 
 A: That is not an exact result, so presumably your professor is using the normal approximation to the binomial to obtain the resulting asymptotic distribution for the squared binomial.  If $X \sim \text{Bin}(n, \theta)$ then for large $n$ we have the approximation $X \sim \text{N}(n\theta, n\theta(1-\theta))$ which then gives:
$$X^2 \sim n \theta (1-\theta) \cdot \text{Chi-Sq} \Bigg( \text{df} = 1, \text{ncp} = \frac{n \theta}{\sqrt{n \theta (1-\theta)}} \Bigg).$$
This result is just a standard application of the central limit theorem, coupled with the continuous mapping theorem.  Note that this is a non-central chi-squared distribution, due to the presence of a non-zero mean for the binomial random variable.  If the binomial test statistic has its mean subtracted prior to squaring then you get a regular chi-squared distribution.
A: *

*First, to address the title, no the square of a binomial random variable isn't chi-squared.

*Now to address the statistic in your comment. The statistic for what you're calling a binomial test (though personally I'd call it a two sample proportions test) doesn't have a binomial distribution; it's a standardized difference in proportion that will be asymptotically standard normal (in sufficiently large samples, the normal distribution is close to the discrete distribution you get from formula). 
(The form you give would correspond to not using a continuity correction.)
At the same time, the statistic for a 2x2 chi-squared test also doesn't have a chi-squared distribution; it's again discrete (though asymptotically chi-squared). 
However, it should be the square of the two-sample proportions test statistic (assuming you also use the uncorrected version). 

*Now to address what I think your professor meant. I expect that your professor was actually referring to a z-test in a one sample proportions test and its correspondence to a goodness of fit chi-squared (and again, the first isn't binomial and the second isn't distributed as chi-squared). 
This correspondence is relatively straightforward to show, you may like to attempt it. 
A: You can derive an asymptotic $\chi^2$-test from the exact binomial test,
but, as stated, the $\chi^2$-distribution is continuous while the
binomial is discrete and result is only asymptotically correct
(approaches the correct result when the sample size grows, under
certain assumptions).
Let's look at a simple binomial test where we have one binary
variable and we would like to evaluate the probability of the observed
frequencies under some null hypothesis. For example, we may have black
and white balls and the variable is the colour of balls. The null
hypothesis is that in the population the probability of white balls is
$q$ and the probability of black balls is $1-q$. You have drawn a sample
of $n$ balls and observed $k$ white balls. Now the exact binomial
probability of obtaining exactly $k$ white balls from $n$ is
$P(K=k|n,q)=bin(n,k)q^k(1-q)^{n-k}$, where $bin(n,k)$ is the binomial
coefficient. If you alternative hypothesis was that the proportion of white balls is more than $q$, the $p$-value is the probability of obtaining at least $k$ white balls, i.e. $P(K\geq k|n,q)$.
Now you can approximate the probability by calculating the
z-score. The mean of the binomial distribution is $nq$ and standard
deviation $\sqrt{nq(1-q)}$. So the z-score is $z=(k-nq)/\sqrt{nq(1-q)}$ and
it follows standard normal distribution. On the the hand, a squared
normally distributed variable (or their sum) follows a $\chi^2$-distribution. So, in this case the square $z^2$ is $\chi^2$-distributed
(with 1 degree of freedom). Note that $z^2$ can be expressed
equivalently as sum $\frac{(k-nq)^2}{nq}+\frac{(n-k-n(1-q))^2}{n(1-q)}$ which is a more familiar expression for the $\chi^2$-statistic.
Note that this $z^2=\chi^2$ does not hold in more complex cases. For
example, if you test independence between two binary variables in a
2x2 table, you can construct two binomial probabilities (assuming that
the margins of one variable are fixed) and obtain two z-scores, $z_1$ and
$z_2$. Now $z_1^2+z_2^2$ follows the $\chi^2$-distribution (with 1 degree of
freedom). But you can derive the same $\chi^2$-test of independence also
from the multinomial model.
