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Let $Y\sim NB(2,\theta)$ represent the trial on which the 2nd success is made.

From a previous question I posed in which many people here helped, I got the UMVUE for $\theta$ as $\frac{1}{Y-1}$. Can someone point me towards the right direction in getting the variance of this estimator? Thanks.

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The Variance is $\mathbb{E}(\left(\frac{1}{Y-1}\right)^2)-\mu^2$. We already know by UMVUE (the UE part) that the expectation of this quantity is $p$ so the only thing you need to find is the second moment. To find this write: $\mathbb{E}[\left(\frac{1}{Y-1}\right)^2]=\sum_{y=2}^{\infty}\left(\frac{1}{y-1}\right)^2{y-1\choose1}p^2(1-p)^{y-2}$. One of the $y-1$s will cancel with the binomial coefficient and you can factor out the constants $p^2$ and $(1-p)^2$ to get: $$\frac{p^2}{(1-p)^2}\sum_{y=2}^{\infty}\left(\frac{1}{y-1}\right)(1-p)^{y}.$$ Now change to a variable $z=y-1$ to get rid of the pesky $y-1$ piece in the denominator. The new equation will be: $$\frac{p^2}{(1-p)^2}\sum_{z=1}^{\infty}\left(\frac{1}{z}\right)(1-p)^{z+1}.$$ Again factor out the $(1-p)$ constant term and note the Taylor series expansion:

$$-\log(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots.$$

Here your $x$ is $1-p$. When you put this in and reduce you should get:

$$\mathbb{E}\left[\left(\frac{1}{Y-1}\right)^2\right] = \frac{-\log(p)\,p^2}{(1-p)}$$

now using the variance formula above you should see that the variance will be:

$$\mathbb{V}ar\left[\frac{1}{y-1}\right]= \frac{-\log(p)\,p^2}{(1-p)} - p^2.$$

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    $\begingroup$ Thanks. I guess there's really no getting around needing to memorize the series expansions. Does this estimator's variance achieve the CRLB? I had computed it before as $\theta^2(1-\theta)/2$ but I am not sure. $\endgroup$ – user164144 Jul 6 '17 at 23:16
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    $\begingroup$ There are very few series one ever has to memorize. The Binomial Theorem $(1+x)^n=\sum_{i=0}^\infty\binom{n}{i}x^i$ covers a great deal of what one encounters in statistics (for $n$ any real number--or, for that matter, a complex number). Integrating it when $n=-1$ gives a series for the log as shown in this answer. From that you can get the inverse tangent. In addition, you need to know $\exp(x)=\sum_{i=0}^\infty x^i/i!$. From this you immediately obtain series for $\sin$, $\cos$, $\sinh$, and $\cosh$; applying it to $x=-y^2/2$ helps with Gaussian functions, and so on. $\endgroup$ – whuber Jul 7 '17 at 0:01
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    $\begingroup$ @whuber Chaining them that way is awesome. Thanks. $\endgroup$ – user164144 Jul 7 '17 at 0:54

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