Suppose \begin{align} f(t)=\alpha g(t)+(1-\alpha) h(t) \end{align} for all $t \in \mathbb{R}$ and for some $\alpha \in (0,1)$.

Suppose that $f(t)$ and $g(t)$ are characteristic function. Does this imply that $h(t)$ is also a chararacteristic function?

Thank you.

  • $\begingroup$ $f$ is a convex combination of $g$ and $h$. So if $g$ and $h$ are characteristic functions, $f$ is also a characteristic functions (consider the mixture distribution). But $h$ is not a convex combination of $f$ and $g$ so $f$ and $g$ are characteristic functions do not imply $h$ is. $\endgroup$ – BGM Feb 2 '17 at 8:13
  • $\begingroup$ @BGM I am not sure that I follow your proof. Any counter examples you have in mind? $\endgroup$ – Boby Feb 2 '17 at 13:06

Firstly recall the valid statement: If $F_1$ and $F_2$ are some distributions with characteristic functions $g(t)$ and $h(t)$, then $F=\alpha F_1+(1-\alpha)F_2$ is their mixture with characteristic function \begin{align} f(t)=\alpha g(t)+(1-\alpha) h(t). \end{align} Take $\alpha=0{,}5$ for simplicity. Then the equality $F=\alpha F_1+(1-\alpha)F_2$ turns to $2F=F_1+F_2$.

If conversely we take some distributions $F$ and $F_1$, the difference $F_2=2F-F_1$ is not obliged to be some distribution. Say, we can take distribution $F$ degenerate at $1$, and $F_1$ degenerate at $0$, and the difference $F_2=2F-F_1$ is not a distribution at all.

Return to characteristic functions. CF for $F$ equals $f(t)=e^{it}$, CF for $F_1$ is $g(t)=1$. The second summand $h(t)$ in the equality $$e^{it}=\frac12 \cdot 1 + \frac12 \cdot h(t)$$ is $h(t)=2e^{it}-1$. This is not CF since the absolute value of this function can be greater than $1$.

  • $\begingroup$ Thanks. Do you know what happens if we restrict the question to real and symmetric characteristic functions? $\endgroup$ – Boby Feb 14 '17 at 14:13
  • $\begingroup$ @Boby Nothing will change. You can find a suitable example. $\endgroup$ – NCh Feb 14 '17 at 15:04

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.