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Let $a=\mathbb E[\log m_0]$ where $\{m_i\}$ are iid and $0<m_i<+\infty $.

Set $$S_n=\sum_{i=0}^{n-1}\left(\log(m_i)-\frac{a}{2}\right).$$

In my course, it's written that $$\mathbb P\{S_n>0\}\leq \mathbb E[e^{\theta S_n}]=\{e^{\frac{-1}{2}\theta a}\mathbb E[m_0^\theta ]\}^n,$$ for all $\theta >0$.

Q1) For the equality, no problem, but why $\mathbb P\{S_n>0\}\leq \mathbb E[e^{\theta S_n}]$ ?

Then, it's written that $$\lim_{\theta\to 0 ^+} (\mathbb E[m_0^\theta ])^{1/\theta }=e^{\mathbb E[\log(m_0)]}=e^a,$$ and thus, there is $\theta >0$ s.t. $$\mathbb E[m_0^\theta ]=(e^{\frac{3}{4}a})^\theta <1.$$

Q2) I neither understand why $\lim_{\theta\to 0 ^+} (\mathbb E[m_0^\theta ])^{1/\theta }=e^{\mathbb E[\log(m_0)]}=e^a,$ not why there is $\theta >0$ s.t. $\mathbb E[m_0^\theta ]=(e^{\frac{3}{4}a})^\theta <1.$

For the limit, I set $$\mathbb E[m_0^\theta ]^\frac{1}{\theta }=e^{\frac{1}{\theta }\log(\mathbb E[m_0^\theta ])},$$ but when but I don't see why it does $e^a$ when $\theta \to 0^+$.

Davide Giraudo
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user352653
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1 Answers1

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Q1 The event $\{S_n\gt 0\}$ is the same as $\left\{e^{\theta S_n}\gt e^{0\theta}\right\}=\left\{e^{\theta S_n}\gt 1\right\}$ and the wanted inequality follows by Markov's inequality.

Q2 The first part follows from this thread; the second from the definition of a limit.

Davide Giraudo
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