Proving $E[2^X] > 2^{\lambda}$ for some RV $X$ ~$Poi(\lambda)$

Question

As the title describes, I've got stuck at solving the sum series, coulden't find any ways to progress:

$E[2^X]=\sum^{\infty}_{i=0} 2^i\cdot P(X=i)$

$=\sum ^{\infty }_{i=0}2^i\cdot e^{-\lambda }\cdot \frac {\lambda ^i}{i!}$

$=e^{-\lambda }\cdot \sum ^{\infty }_{i=0}\frac {(2\lambda )^i}{i!}$

EDIT: After Kazuki Okamura edited my question to change the upper bounds of the sum from $\lambda $ to $\infty $ I'd like to ask, why so?

Is this because $\lambda $ which represents the number of successes per time unit can be infinite?

I cant seem to wrap my head around the idea of limiting $\lambda $.

Answer 1

We say that a random variable $X$ follows the Poisson distribution with the parameter $\lambda > 0$ if $P(X = i) = \exp(-\lambda) \frac{\lambda^i}{i!}$ for every integer $i \ge 0$. Since $\sum_{I \ge 0} P(X=i) = 1$, this is well-defined.

Let $a > 0$. Then, $$ E[a^X] = e^{-\lambda}\sum_{i=0}^{\infty} \frac{(a\lambda)^i}{i!} = \exp((a-1)\lambda). $$ Since $1+x \le e^x$ for every $x \in \mathbb R$, $a - 1 \ge \log a$. Hence, $E[a^X] \ge a^{\lambda}$. In particular, we can apply this to the case that $a=2$.

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If $a \ne 1$, then, $a-1- \log a > 0$ and hence, $$ \frac{E[a^{X}]}{a^{\lambda}} = \exp(\lambda(a-1- \log a)) > 1.$$

Answer 2

Jensen's inequality (for expected values) states that for a convex function $f \colon \mathbb R \to \mathbb R$ we have $$\mathbb{E}[f(X)] \ge f(\mathbb{E}[X])$$ and $>$ if $f$ is strictly convex. As the exponential function $f(x) := 2^x$ (as noted im the other answer, the 2 can be replaced with any other positive number $a > 0$) is strictly convex and for $X \sim \text{Poiss}(\lambda)$ we have $\mathbb{E}[X] = \lambda$, Jensen's inequality yields $$\mathbb{E}[2^X] > 2^{\mathbb{E}[X]} = 2^{\lambda}.$$