Let $\phi_1,\ldots,\phi_n$ denote characteristic functions for distributions on the real line. Let $a_1,\ldots,a_n$ denote nonnegative constants such that $a_1+\ldots+a_n = 1$. Show that

$$\hspace{20mm} \phi(t) = \sum_{j=1}^{n}a_j\phi_j(t), \hspace{8mm}-\infty<t<\infty$$


Let $a = 1$.

$$\phi(t) = \mathrm{E}\left[\exp(aitX)\right] = \int_{-\infty}^{\infty} \exp(aitX)\times p(x)\,\mathrm{d}x $$

I'm not sure how to proceed. I would assume that the additive property of the characteristic functions has something to do with the exponential components.

  • 2
    $\begingroup$ Since you have not stated what $\phi(t)$ is, your first displayed equation is, in effect, a definition of $\phi(t)$, and there is nothing to show. Have you been asked to prove that $\phi(t)$ as defined above is a valid characteristic function corresponding to the distribution of some random variable? If so, think of mixture distributions. $\endgroup$ Jun 27, 2014 at 10:41

1 Answer 1


I suppose that you want to show that $\phi(t)$ is characteristic function. For that define a new variable $N$ with distribution given with the convex coefficients: $P(N=i)=a_i.$ Further suppose that $X_i$ is a random variable with $\phi_i(t)$ and define $$Y=\sum_{j=1}^nX_j I{[N=j]}, \text{ where } I{[N=j]}=1 \text{ if } N=j, 0 \text{ otherwise}$$ The characteristic function of $Y$ is $\phi(t)$, because $$E\exp(itY)=\sum_{j=1}^nE\{\exp(itX_j)P({[N=j]})\}=\sum_{j=1}^n\phi_j(t)a_j=\phi(t)$$

  • $\begingroup$ If the n is going to infinite, this conclusion is still valid? $\endgroup$
    – user58543
    Oct 13, 2014 at 16:00
  • $\begingroup$ Yes, it is still valid. The crucial point is if you can switch E (integral) and sum. The answer is yes, because we can use en.wikipedia.org/wiki/Dominated_convergence_theorem. The hardest is to find dominating function. But we know that $$E\sum_{j=1}^{\infty}\exp(itX_j)a_j=\int_{-\infty}^{\infty}\sum_{j=1}^{\infty} \exp(itx)a_jf_j(x)d\mu(x)=\int_{-\infty}^{\infty}s(x)d\mu(x)$$ And dominating variant can be found in the following way $$|s(x)|\le\sum_{j=1}^{\infty}|\exp(itx)a_jf_j(x)|\le max_j f_j(x)=g(x).$$ In the end we know that $\int_{-\infty}^{\infty}g(x)d\mu(x)=1.$ $\endgroup$
    – Fimba
    Oct 13, 2014 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.