Characteristic function and third moment

Question

Examining a problem I find that: $E[(Y_n)^3] = \sum_{i=1}^nE[(X_i)^3]$. Where $Y_n=\sum_{i=1}^nX_i$.

Now, I'm not able to understanding the following step: $$\phi^{'''}_{Y_n}(0) = - i\sum_{i=1}^nE[(X_i)^3]$$ where $i=\sqrt{-1}$

I think that they applied the following formula: $E[(Y_n)^3] = (-i)^3\phi^{(3)}_{Y_n}$. And then they move the "i" in the other side. But I'm not able to understand how. Because in my computation I have to multiply in both side for "i" at the power of something, therefore I never have $i^1$.

Answer 1

They did exactly as you said but simplified it: $\frac{1}{(-i)^3} = -i$.

Reason: \begin{align*} \frac{1}{(-i)^3} &= \frac{1}{(-1)^3i^2 i}\\ &=\frac{1}{(-1)(-1)i}\\ &= \frac{1}{i}\\ &= -i. \end{align*} The reason that $\frac{1}{i}=-i$ is because $i\cdot (-i) = -i^2 = 1$.

Answer 2

Recall the definition of the characteristic function $$ \phi_X(t)=\mathbb Ee^{itX}. $$ Thus, if we take derivatives with respect to $t$ inside the expectation we have $$ \phi_X^{(n)}(t)=\mathbb E(iX)^ne^{itX}. $$ Now set $t=0$ so that $$ \phi_X^{(n)}(0)=\mathbb E(iX)^n. $$ I think your question is answered by the case $n=3$.