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Let $X_1, ..., X_n$ i.i.d. from $X$ where $X$ is binary and we have p = P{$X=1$} ,$p\in (0,1)$. Let $\alpha = \frac{p}{1-p}$. Define $\bar{a} = \frac{\bar{X}}{1-\bar{X}}$, show that $\bar{\alpha}$ is unbiased and consistent for $\alpha$.

So to prove unbiased I show: $E[\bar{X}] = E[X]$ since we have $X$ binary, $p = E[X]$. By the continuous mapping theorem since the numerator $=p$ in probability and the denominator $=1-p$ we have unbiased $\bar{\alpha}$.

To prove consistency, we note $\bar{X} \rightarrow E[X]$ (in probability) by weak law of large numbers and the same for $1-\bar{X}$. By continuous mapping, $\alpha$ is consistent.

I feel like there there are flaws in my arguments somewhere.

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The problem is that $\bar\alpha$ is not even defined on the whole probability space (as soon as $p\ne0$) since $[\bar X=1]$ has positive probability $p^n$. In particular $E(\bar\alpha)=+\infty$. (And you might want to review the statement of the continuous mapping theorem.) – Did Nov 5 '12 at 15:26
So if for random variables X and Y $X\rightarrow a$ and $Y\rightarrow b$ and $g$ is continuous at $(a,b)$ then $g(X, Y) \rightarrow g(a,b)$. That means I should try to show individually that $1-\bar{X} \rightarrow 1-p$? – Emir Nov 5 '12 at 15:36
Because $\bar\alpha=+\infty$ on $A=[\bar X=1]$ and $P(A)\ne0$. – Did Nov 5 '12 at 15:56
And? Note that $[\bar X=1]=[X_1=X_2=\cdots=X_n=1]$. – Did Nov 5 '12 at 15:58
$\Pr(\bar{X}=1)=p^n \gt 0$ so $\Pr(\bar{\alpha}=\infty)\gt 0$ so $E[\bar{\alpha}]=\infty$. – Henry Nov 5 '12 at 16:22

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