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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

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

Attempt:

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.

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    $\begingroup$ 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

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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 $$E\exp(itY)=\sum_{j=1}^nE\{\exp(itX_j)P({[N=j]})\}=\sum_{j=1}^n\phi_j(t)a_j=\phi(t)$$

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  • $\begingroup$ 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

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