# New characteristic functions from old

I am doing an exercise which says: If $f$ is a characteristic function, then show that $$F(t):= \int_0^{\infty} f(ut)e^{-u}du$$ is again a characteristic function.

Is this answer correct? Let $Z$ be an exponentially distributed random variable with parameter $1$. Let $X$ be a random variable with characteristic function $f$. Then

\begin{align} F(t) &= \mathbb{E} \left[ \left( \mathbb{E} e^{itsX} \right) \Bigg{\vert}_{s=Z}\right] \\ &= \mathbb{E} \left[ \mathbb{E}(e^{itZX} \big{\vert} Z ) \right]\\ &=\mathbb{E}(e^{itZX}) \end{align}

where it is the second equality that uses the independence. This shows that $F$ is the characteristic function of the random variable $ZX$.

Bochner's Theorem: A function $F:\mathbb R\to \mathbb C$ is the characteristic function for some random variable if, and only if, $F$ is positive-definite, continuous at the origin and $F(0)=1$.

Positive definiteness: The function $f$, being a characteristic function, is positive-definite. Consider, for $\ r,s=1,\ldots,n,\$ the arbitrary numbers $\ \xi_r, \bar{\xi_s}\in\mathbb C,\ t_r, t_s\in\mathbb R$. Observe that

$$\begin{array}{crl}\sum\limits_{r,s=1}^{n}\xi_r\bar{\xi_s}F(t_r-t_s) = \sum\limits_{r,s=1}^{n}\xi_r\bar{\xi_s}\int\limits_{0}^{\infty}f(ut_r-ut_s)e^{-u}\text du \\= \int\limits_{0}^{\infty}\sum\limits_{r,s=1}^{n}\xi_r\bar{\xi_s}f(ut_r-ut_s)e^{-u}\text du \geqslant 0\,, \end{array}$$

which shows that $F$ is positive-definite.

Continuity at the origin: Clearly, $F(0)=1$. Now, for the sequence $\frac{1}{n}$, we see that $\lim\limits_{n\to\infty}|f(u\frac{1}{n})-1|=0$ and $\ |f(u\frac{1}{n})-1|\leqslant 2\$ for all $n, u>0\$, by the continuity and boundedness of $f$. Therefore, by Lebesgue dominated convergence,

$$\begin{array}{ccl}\lim\limits_{n\to\infty}|F(\frac{1}{n})-F(0)| &=& \lim\limits_{n\to\infty}|\int_{0}^{\infty}(f(u\frac{1}{n})-1)e^{-u}~\text du| \leqslant \lim\limits_{n\to\infty}\int_{0}^{\infty}|f(u\frac{1}{n})-1|e^{-u}~\text du \\ &=& \int_{0}^{\infty}\lim\limits_{n\to\infty}|f(u\frac{1}{n})-1|e^{-u}~\text du = 0\,,\end{array}$$

so $F(t)$ is continuous at $0$.

We conclude that $F$ is the characteristic function for some random variable.

• This is certainly one way of doing it, but I think Bochner's theorem for this example is avoidably messy. My proposed solution (which I am now confident is correct) interprets the new characteristic function simply as that of a random variable multiplied by an (independent) random number in $(0,1)$. Nonetheless, thank you for your answer. – Frank Apr 8 '15 at 14:12
• @Frank, Indeed, your solution is elegant (I should have said this)! In fact, upon reading it the first time, it seemed too clever to me :). For, it implied all of the properties in Bochner's theorem hold. So, naturally, one could ask just how general the approach is? Is it sufficient that we consider $\int_{0}^{\infty}f(ut)g(u)du$ where $\int_{0}^{\infty}g(u)du = 1$? So, partly to put "Bochner's" out there for those who might benefit from seeing it in action, and partly as a sanity check, I put up this alternative solution which hints at the answer to my follow up question being yes. – ki3i Apr 8 '15 at 15:54
• @Frank,...although, I cannot see how one might use your kind of existence argument (i.e. identifying a suitable independent random variable) to see the result in this generalisation of your problem. Food for thought, I guess. – ki3i Apr 8 '15 at 16:04