Let $X\sim \mathcal{N}(0,I_p)$ and $\tau=\sqrt{(2-\varepsilon)\log p}$ and $\varepsilon>0$. I want to prove that for sufficiently small $\varepsilon>0$ the following holds:
$$ \mathbb{E}\left[ \frac{e^{2 \tau X_1}}{\sum_{i=1}^p e^{\tau X_i}} \right]=o(1), \quad p \rightarrow \infty. $$ Can anyone provide useful insights about how to show this?
My attempt: $\mathbb{E}\left[ \frac{e^{2 \tau X_1}}{\sum_{i=1}^p e^{\tau X_i}} \right]=\mathbb{E}\left[ \frac{1}{e^{-\tau X_1}+\sum_{i=2}^p e^{\tau(X_i-2X_1)}} \right] \leq \mathbb{E} \left[ \frac{1}{e^{-\tau X_1} +e^{\tau(X_{max}-X_1)} } \right]$, where $X_{max}=\max_{2 \leq i \leq p}X_i$.
We know that $X_{max}$ roughly behaves like $\sqrt{2\log p}$ and $X_1$ behaves like a constant. But how do I make this argument rigorous and prove that the expectation is $o(1)$?