For $Y \sim Beta(n, 1)$, showing that $\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx$ converges in probability to $-\infty$?

Question

Suppose that $Y \sim Beta(n, 1)$. I am interested in showing that

$$ \frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx $$

converges in probability to $-\infty$. (I was told in a book this is true)

My set-up is to show for $K>0$:

$$ \lim_{n \to \infty}P\left(\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx <-K\right) = 1 $$

Now,

\begin{align} P\left(\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx <-K\right) &= P\left(\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx 1-e^{-nK}\right) \\ &= 1-\left(1-e^{-nK}\right)^n \end{align}

Above, my 2nd to 3rd inequality was obtained as $x^{-n}e^{-(x-1)^{-2}}$ is bounded below by $x^{-1}e^{-(x-1)^{-2}}$, hence the terms inside the probability for $x^{-n}e^{-(x-1)^{-2}}$ is a subset of the terms inside the probability for $x^{-1}e^{-(x-1)^{-2}}$.

The same for my 3rd to 4th inequality.

Applying the limit, the entire probability term goes to $0$. Hence:

$$ \lim_{n \to \infty}P\left(\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx <-K\right) = 0 $$

which is different from the result above. Does anyone know where I might have messed up? Thanks.

Answer 1

Let $\{Y_n\}_{n\geq1}$ a sequence of random variables defined on the same probability space such that $Y_n\sim Beta(n,1)$ and $Y_n

The problem in your demonstration is, as stated by NCh, that $\begin{align}P\left(\int_{Y}^{1}x^{-1}e^{-(x-1)^{-2}}dx

Proof of $\frac{1}{n}log\int_{\frac{n}{n+1}}^1 x^{-n}e^{-(x-1)^{-2}}dx\to-\infty$:

Using L'hôpital, if $\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx \to 0$ $$\begin{align*} \lim_{n\to\infty}\frac{log\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx}{n} &=\lim_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^{n}e^{-\left(n+1\right)^{2}}/\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx}{1}\\ &=\lim_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^{n}e^{-\left(n+1\right)^{2}}}{\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx}\\ &=e\lim_{n\to\infty}\frac{-2\left(n+1\right)e^{-\left(n+1\right)^{2}}}{\left(1+\frac{1}{n}\right)^{n}e^{-(n+1)^{2}}}\\ &=\lim_{n\to\infty}-2\left(n+1\right)\\ &=-\infty\end{align*} $$

Proof of $\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx \to 0$:

$\begin{align*}\lim_{n\to\infty}\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx &\leq\lim_{n\to\infty}\int_{\frac{n}{n+1}}^{1}\left(\frac{n}{n+1}\right)^{-n}e^{-(\frac{n}{n+1}-1)^{-2}}dx\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{-n}e^{-(\frac{n}{n+1}-1)^{-2}}\left(1-\frac{n}{n+1}\right)\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}e^{-(n+1)^{2}}\left(\frac{1}{n+1}\right)\\ &=0\end{align*}$

Proof of $g(u)=\frac{1}{n}log\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx$ being concave:

let f,g be concave twice differentiable functions, if f is increasing then $$f\left(h\left(u\right)\right)'' =f''\left(h\left(u\right)\right)h'\left(u\right)^{2}+f'\left(h\left(u\right)\right)h''\left(u\right)\leq0$$ if we apply that first to $f(u)=u^{\frac{1}{n}}$, $h(u)=\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx$

and then to $f(u)=log(u)$ and $h(u)=\left(\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx\right)^{\frac{1}{n}}$

Proof of $h(u)=\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx$ being concave: $$ \begin{align*} h'(u) & =u^{-n}e^{-(u-1)^{-2}}\\ h''(u) & =-nu^{-n}e^{-(u-1)^{-2}}-2u^{-n}e^{-(u-1)^{-2}}(1-u)^{-3} \end{align*} $$