0
$\begingroup$

In probability theory, the characteristic function of a random variable $X$ is defined as

$$ \varphi_X(t) =\mathrm{E}\left(e^{\mathrm{i}tX}\right), $$

We further know that the characteristic function of the sum of $N$ independent random variables, $Z=\sum_{k=1}^{N}X_k$, is simply $$ \varphi_Z(t)=\mathrm{E}\left(e^{itZ}\right)=\prod_{k=1}^N\phi_{X_k}(t). $$

Thus, the product of characteristic functions has a clear interpretation in terms of the transformation of the sum of independent random variables.

For quite some time now, I am wondering whether the sum of characteristic functions may also have an interpretation in terms of random variables and their probability distributions?

For example, we know that the sum of an exponential and a standard normal random variable has the transform

$$ \varphi_{Exp(\lambda)+N(\mu,\sigma)}(t)=\frac{1}{1-i\lambda t}e^{i\mu t - \frac{1}{2}t^2\sigma^2} $$

But is there a meaningful interpretation to a characteristic function of the form

$$ \varphi_{?}(t)=\frac{1}{1-i\lambda t}\mathbf{+}e^{i\mu t - \frac{1}{2}t^2\sigma^2} $$

$\endgroup$
1
  • 2
    $\begingroup$ Never encountered and I would be quite surprised if there is a meaningful interpretation of it. No reason to believe so. $\endgroup$
    – drhab
    Dec 9, 2017 at 8:45

2 Answers 2

3
$\begingroup$

The definition of a characteristic function $$\varphi_X(t) =\mathrm{E}\left(e^{\mathrm{i}tX}\right),$$ implies $$\varphi_X(0) =\mathrm{E}1=1,$$ so $\varphi_X(t)+\varphi_Y(t)$ can't be a characteristic function. Of course, we can normalize it: $$\varphi_Z(t)=\frac12(\varphi_X(t)+\varphi_Y(t))=\frac12\mathrm{E}\left(e^{\mathrm{i}tX}\right)+\frac12\mathrm{E}\left(e^{\mathrm{i}tY}\right).$$ The probabilistic interpretation is obvious: you toss a fair coin, and "head" means $Z=X$, "tail" is $Z=Y$.

$\endgroup$
1
  • $\begingroup$ Yes, that makes very much sense to: A weighted sum of char. functions is simply the characteristic function inversion of a mixed distribution! Thank you very much. $\endgroup$
    – MHeld
    Dec 11, 2017 at 6:31
0
$\begingroup$

The sum of two characteristic function cannot be a characteristic function itself… a characteristic function fullfils $\varphi(0) = 1$ and is bounded by $1$ so for the sum of two characteristic functions it holds $$\varphi_{X_1}(0) + \varphi_{X_2}(0) = 2$$ Hence two main characteristics are violated…

$\endgroup$
2
  • $\begingroup$ Yes, you are absolutely right -- and I was a bit sloppy. But after rescaling by $1/\sum \phi(0)=1/N$, The result of @Professor-Vector seems worthwile. $\endgroup$
    – MHeld
    Dec 11, 2017 at 6:36
  • $\begingroup$ That's why I already upvoted :-) $\endgroup$
    – Gono
    Dec 11, 2017 at 8:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .