# Convergence of error term

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.

• 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 $n \rightarrow \infty$. – AdamO Apr 22 at 14:26
• @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... – mkmlp Apr 22 at 14:36
• Then look for a bounding function and make it easier on yourself – AdamO Apr 22 at 15:33

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?

• $\exp(1)+\exp(-1)$! Wow this transformation is neat! Thanks a million! – mkmlp Apr 22 at 21:01
• @mkmlp You're welcome. I guess this means that the expression does go to zero (due to logarithms)? – Alecos Papadopoulos Apr 22 at 21:16
• 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 – mkmlp Apr 22 at 21:20