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I was deriving the ratio of a Laplace approximation with the true quantity and I got this:

$$ \left(n\bar{x} + \alpha - \frac{1}{2}\right)\log\left(\frac{\bar{x}+(a-1)/n}{\bar{x}+\alpha/n}\right) + n\log\left(\frac{1+(\alpha+\beta-1)/n}{1+(\alpha+\beta-2)/n}\right), $$

where $\bar{x}$ is the sample mean of an i.i.d. Binomial($1$, $\theta$) random sample of size $n$, and $\alpha,\beta>0$.

I'm having trouble seeing if this thing converges to $0$. Can anyone please help? A gentle hint would be greatly appreciated.

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  • $\begingroup$ Two popular ways of showing the limiting quantity of a function in statistics is: a) use L'Hospital's Rule and b) find a bounding function above or below who's limit is 0 or $\infty$ respectively. It seems pretty straightforward albeit detailed to do a), my hunch is that the quantity, as written, does not go to 0 as $n \rightarrow \infty$. $\endgroup$ – AdamO Apr 22 at 14:26
  • $\begingroup$ @AdamO Thanks for the help. Although, I don't think L'Hospital directly applies here since (after absorbing the multiplicative term into the log) the numerator and the denominator don't go to $0$ or $\infty$ simultaneously. I might be wrong... $\endgroup$ – mkmlp Apr 22 at 14:36
  • $\begingroup$ Then look for a bounding function and make it easier on yourself $\endgroup$ – AdamO Apr 22 at 15:33
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Set

$$z_n \equiv n\bar x +\alpha, \;\;\; \alpha + \beta -2 \equiv \gamma,\;\;\; x_n \equiv n+\gamma$$

The the expression can be re-written as

$$(z_n - 1/2)\cdot\ln\left(\frac{z_n -1}{z_n}\right) + (x_n -\gamma) \cdot \ln\left(\frac{x_n +1}{x_n}\right) $$

$$=\ln\left(1-\frac{1}{z_n}\right)^{z_n} - \ln\left(1-\frac{1}{z_n}\right)^{1/2} + \ln\left(1+\frac{1}{x_n}\right)^{x_n} - \ln\left(1+\frac{1}{x_n}\right)^{\gamma} $$

Do you see where this approach is going?

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  • $\begingroup$ $\exp(1)+\exp(-1)$! Wow this transformation is neat! Thanks a million! $\endgroup$ – mkmlp Apr 22 at 21:01
  • $\begingroup$ @mkmlp You're welcome. I guess this means that the expression does go to zero (due to logarithms)? $\endgroup$ – Alecos Papadopoulos Apr 22 at 21:16
  • $\begingroup$ yes you’re right. Sorry when I wrote $\exp(1)$ and $\exp(-1)$ I meant those are the limits of the terms inside the logarithms...got too excited $\endgroup$ – mkmlp Apr 22 at 21:20

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