In a sum of high-variance lognormals, what fraction comes from the first term?

Question

If $X_i \overset{\textrm{iid}}{\sim} \text{Lognormal}(0, \sigma^2)$ for $i=1,\ldots,n$ and $Y_1 = X_1 / \sum_{j=1}^n X_j$, then I would expect that a particular* limiting distribution of $Y_1$, namely $ \lim_{\sigma \to \infty} Y_1$, would be very simple: $1$ with probability $1/n$ and $0$ with probability $1-1/n$. Is there an (easy) way to prove this?

*NB: I'm interested in the limit with respect to $\sigma$, not $n$. $n$ is not necessarily large.

Answer 1

Let's start with $Z_i \overset{\textrm{iid}}{\sim} \text{Normal}(0, 1)$ and let $X_i=e^{\sigma Z_i}=\left(e^{ Z_i}\right)^\sigma$ and $Y_i = X_i / \sum_{j=1}^n X_j$.

With probability $\frac1n$, $Z_1= Z_{(n)} > Z_{(n-1)}$ in which case

• $Y_1 = Y_{(n)}\ge \frac{\left(e^{ Z_{(n)}}\right)^\sigma}{\left(e^{ Z_{(n)}}\right)^\sigma+(n-1)\left(e^{ Z_{(n-1)}}\right)^\sigma} =\dfrac{1}{1+(n-1)\left(e^{ Z_{(n-1)}-Z_{(n)}}\right)^\sigma} \to 1$ as $\sigma\to \infty$.

Similarly, with probability $1-\frac1n$, $Z_1\le Z_{(n-1)} < Z_{(n)}$ in which case

• $Y_1 \le 1-Y_{(n)} \to 0$ as $\sigma\to \infty$.