9
$\begingroup$

Suppose that $X$ ~ $Binomial(n,p)$ for $0 < p < 1$

Why does no unbiased estimator exist for $1/p$?

My approach:

We try to find the structure of $E_p(U(x))$, where $U(x)$ is any estimator of $1/p$.

Now, we will have:

$\sum{U(x)\binom{n}{x}p^x(1-p)^{n-x}}<\sum{U(x)\binom{n}{x}}=M(n)<\infty$

so that the expectation is bounded above. So this is supposed to mean that if $p < 1/M(n)$, then the expectation cannot attain $p$ but I am not sure why the above argument even makes sense and what being bounded means for the expectation.

$\endgroup$
3
  • $\begingroup$ What does $M(n)$ represent here? Just some bounded function of $n$? Or something more specific? Why did you switch from $p$ to $\theta$? $\endgroup$ – Glen_b Feb 19 '14 at 5:37
  • 1
    $\begingroup$ It's still not clear to me what $M(n)$ is. $\endgroup$ – Glen_b Feb 19 '14 at 5:51
  • 3
    $\begingroup$ Hint: Using the definition, the expectation is a polynomial (in $p$!) no matter what $U$ is. This polynomial is supposed to be equal to a rational function with numerator degree of zero. Conclusion? $\endgroup$ – cardinal Feb 19 '14 at 13:06
5
$\begingroup$

By definition, an estimator of any property of the distribution of $X$ is a function $t$ of the possible values of $X,$ here equal to $0, 1, \ldots, n.$

Given $n,$ assume $X$ has some Binomial$(n,p)$ distribution where $p$ is known only to lie within a given set $\Omega \subset[0,1].$ We will say more about $\Omega$ at the end.

The expectation of $t$ is its probability-weighted average,

$$E[t(X)] = \sum_{x=0}^n \Pr(X=x) t(x) = \sum_{x=0}^n \binom{n}{x}p^x(1-p)^{n-x} t(x).$$

(Note that the expressions "$t(x)$" are just numbers, one for each $x=0,1,\ldots, n.$)

This expectation depends on $p.$ In the specific case where $1/p$ is to be estimated, the estimator is unbiased when it equals $1/p$ for all values of $p \in\Omega;$ that is,

$$\frac{1}{p} = E[t(X)] = \sum_{x=0}^n \binom{n}{x}p^x(1-p)^{n-x} t(x).\tag{*}$$

Since $p\ne 0,$ this is algebraically equivalent to

$$\eqalign{ 0 &= pE[t(x)] - 1 \\ &= -1 + \sum_{x=0}^n \binom{n}{x}p^{x+1}(1-p)^{n-x} t(x)\\ &= -1 + \sum_{x=0}^n \binom{n}{x}p^{x+1}\sum_{i=0}^{n-x}\binom{n-x}{i}(-p)^i t(x)\\ &= -1 + \sum_{x=0}^n\sum_{i=0}^{n-x}(-1)^i t(x) \binom{n}{x}\binom{n-x}{i}\,p^{x+1+i}\\ &= -1 + \sum_{k=1}^{n+1} \left(\sum_{i=0}^{k-1}(-1)^i t(k-1-i)\binom{n}{k-1-i}\binom{n-k+1+i}{i}\right)\, p^k\\ &= \sum_{k=0}^{n+1} a_k\, p^k }$$

where $a_0=-1$ and

$$a_k = \sum_{i=0}^{k-1}(-1)^i t(k-1-i)\binom{n}{k-1-i}\binom{n-k+1+i}{i}$$

are constants determined by $t.$

This is explicitly a nonzero polynomial of degree at most $n+1$ in $p$ and therefore can have at most $n+1$ zeros. If, then, $\Omega$ contains more than $n+2$ values, this equation cannot hold for all of them, whence $t$ cannot be unbiased.

Generalizations of this result to certain other functions of $p,$ besides $1/p,$ should be obvious.


The reason why this argument does not generalize to, say, estimating $p,$ is that similar calculations give a polynomial whose coefficients actually can be reduced to zero by a suitable choice of $t:$ that's why it was crucial to observe that the polynomial determined by $a_0, a_1, \ldots, a_{n+1}$ is nonzero (because $a_0=-1$ no matter what).

Take, for instance, the case $n=2.$ The condition $(*)$ of unbiasedness becomes, for all $p\in\Omega,$

$$\eqalign{ 0 &= -p + E[t(X)] \\&= -p + \left[t(0)(1-p)^2 + 2t(1)p(1-p) + t(2)p^2\right] \\ &= t(0) + (-1-2t_0+2t_1)p + (t(0)-2t(1) + t(2))p^2. }$$

Working from left to right we find that the coefficients can all be made zero by setting $t(0)=1,$ then $t(1)=1/2,$ and finally $t(2) = 1.$ This is the only set of choices that does so. Thus, when $n=2$ and $\Omega$ contains at least three elements, this estimator $t$ is the unique unbiased estimator of $p.$


Finally, as an example of why the content of $\Omega$ matters, suppose $\Omega=\{1/3, 2/3\}.$ That is, we know $X$ counts the heads in two flips of a coin that favors either tails or heads by odds of $2:1$ (but we don't know which way). An unbiased estimate $1/p$ is obtained by the estimator $t(0) = 11/2,$ $t(1) = 1 = t(2).$ The check is straightforward: when $p=1/3$, the expectation of $t$ is

$$(2/3)^2\,t(0) + 2(2/3)(1/3)\,t(1) + (1/3)^2\,t(2) = (4/9)(11/2) + 4/9 + 1/9 = 3$$

and when $p=2/3$ the expectation is

$$(1/3)^2\,t(0) + 2(1/3)(2/3)\,t(1) + (2/3)^2\,t(2) = (1/9)(11/2) + 4/9 + 4/9 = 3/2.$$

In each case the expectation indeed is $1/p.$ (It is amusing that none of the values of $t,$ though, are actually equal $3$ or $3/2,$ which are the only two possible values of $1/p.$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.