# Characteristic functions with the same derivatives

Is there an example of two characteristic functions having the same derivative?

More concretely: Suppose that $\phi_X(t)$ and $\phi_V(t)$ are two characteristic functions.

Q1: Does there exists $\phi_X(t) \neq \phi_V(t)$ such that $\phi_X^\prime(t) = \phi_V^\prime(t)$?

Q2: If there are such $\phi_X(t)$ and $\phi_V(t)$, can we come up with sufficient condition on characteristics functions such that the derivatives are unique?

My attempt: Suppose that $\phi_X^\prime(t) = \phi_V^\prime(t)$ then by the fundamental theorem of calculus: \begin{align} \phi_X(t)-\phi_X(0)&= \int_0^t \phi_X^\prime(u) du\\ &=\int_0^t \phi_V^\prime(u) du\\ &= \phi_V(t)-\phi_V(0) \end{align}

and since $\phi_V(0)=\phi_X(0)=1$ we have that $\phi_X(t)=\phi_V(t)$. So, it seems that the derivative must also be unique. Is this a rigorous argument?

Note that your argument uses essentially no special facts about characteristic functions. We could generalize this somewhat, to get the following:

If $f'(x) = g'(x)$ on some compact interval $[a,b]$, both $f'(x)$ and $g'(x)$ are continuous, and $f(0) = g(0) = 1$, then $f(x) = g(x)$ on $[a,b]$.

Here, we have that $f',g'$ are continuous on $[a,b]$ as that's required to apply the fundamental theorem of calculus.

You certainly proved that this is true. The question is then if our particular situation matches this general situation. So, the question reduces to "Does the characteristic function of a random variable always have a continuous first derivative?".

Let's look! Let $p_X(x)$ be the pdf of $X$. Define: $$\varphi_X(t) = \mathbb{E}[e^{itX}] = \int_{-\infty}^\infty e^{itx}\rho_X(x)\mathrm{ d}x$$ The first derivative of this is: $$\frac{\partial}{\partial t}\varphi_X(t) = \frac{\partial}{\partial t}\int_{-\infty}^\infty e^{itx}\rho_X(x)\mathrm{ d}x = \int_{-\infty}^\infty ix e^{itx}\rho_X(x)\mathrm{ d}x$$ Note that in particular we have that $\varphi_X'(0) = i\mathbb{E}[x]$.

One issue that can appear is if $\mathbb{E}[x] = \infty$ (this happens sometimes, see for example the Pareto distribution).

If we know that $X$ is such that $\int_{-\infty}^\infty ixe^{itx}\rho_X(x)\mathrm{ d}x = \mathbb{E}[X e^{itX}] <\infty,\quad \forall t$, then your proof should go through! Otherwise, the assumption fails, so the proof will be insufficient (even if the end result ends up being correct --- you'd just need another way to prove it!).

• Thanks. How would you generalize this to the characteristic function of random vectors when we replace derivative by the gradient. What would you have to you instead of the fundamental theorem of calculus? – Boby May 9 '18 at 17:14
• Writing $E[Y]<\infty$ when $Y$ is complex-valued... add $\lvert\cdot\rvert$? – Clement C. May 9 '18 at 20:30