Exercise 8.4.3

Question

\begin{enumerate}[label=(\alph*)] 
\item Use the results of Exercise 2.8.7 and the binomial formula to show that $E(x+y) = E(x)E(y)$ for all $x,y\in \mathbf{R}$.
\item Show that $E(0) = 1$ , $E(-x) = 1/E(x)$, and $E(x) > 0$ for all $x \in \mathbf{R}$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item
$$\begin{aligned}
    E(x+y) &= \sum^\infty_{n=0} \frac{(x+y)^n}{n!} = \sum^\infty_{n=0} \frac{1}{n!} \sum^n_{i=0} \frac{n!}{i!(n-i)!} x^i y^{n-i} = \sum^\infty_{n=0} \sum^n_{i=0} \left(\frac{x^i}{i!}\right) \left(\frac{y^{n-i}}{(n-i)!}\right) \
    &= \sum^\infty_{i=0} \sum^\infty_{j=0} \left(\frac{x^i}{i!}\right) \left(\frac{y^j}{j!}\right) = \left(\sum^\infty_{i=0} \frac{x^i}{i!}\right)\left(\sum^\infty_{i=0} \frac{y^i}{i!}\right) = E(x) E(y)
\end{aligned}
    $$
\item For $E(0)$, all terms for $n \geq 1$ become 0, so $E(0) = 1$. We showed somewhat informally in Exercise 6.6.5(c) that $E(-x) E(x) = 1$ by collecting common terms. However, Exercise 2.8.7 lets us conclude that $E(-x)E(x)$ does in fact equal $\sum^\infty_{n=0} d_n$ where $d_n$ is defined as in Exercise 6.6.5(c). Finally, it's clear that $E(x) > 0$  for $x \geq 0$ simply because all terms are positive; then $E(-x) = 1/E(x) > 0$ and so $E(x) > 0$ for $x < 0$ as well.
 \end{enumerate}

Exercise 8.4.3

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Exercise 8.4.3

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