Finding the derivative of a characteristic function

Problem summary​

I have the characteristic function of a probability distribution but I'm having difficulty obtaining its derivative.

Background​

I am reading the following paper: Schwartz, Lowell M. (1980). On round-off error. Analytical Chemistry, 52(7), 1141-1147. DOI:10.1021/ac50057a033.

The early part of the paper is concerned with deriving the characteristic function for a distribution that relates to ##n## quantized (rounded) samples drawn from a Gaussian distribution with mean, ##\mu##, standard-deviation ##\sigma##, where the width of the quantization interval is##q##. The characteristic function is stated as being:
$$
G_{\bar{y}}(t) = \left(\sum_{k=-\infty}^{\infty} \frac{\sin \pi \left( \frac{t}{n \phi} + k \right)} {\pi \left( \frac{t}{n \phi} + k \right)} \:\, \exp\left[ -\frac{1}{2}(\frac{t}{n} + k \phi)^2 \sigma^2 + i \, \mu (\frac{t}{n} + k \phi) \right]\right)^n,
$$
where ##\phi=2\pi/q##.

Up to that point in the paper, I'm OK. I can follow Schwartz's argument, and do my own derivation of his results, including the derivation of ##G_{\bar{y}}(t)##.
Beyond that, however, I'm stuck.

After presenting the characteristic function, Schwartz states that "the first moment is found after lengthy but straightforward manipulations" to be
$$
M_1 = E(\bar{y}) = \mu + \frac{q}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \, X \, \sin\left( \frac{2 \pi k \mu}{q} \right),
$$
with ##X= \exp(-2 k^2 {\pi}^2 {\sigma}^2/q^2)## ... this being how Schwartz himself present the result). In calculating ##M_1##, Schwartz relies on the fact that the ##r##-th moment (if it exists) for a PDF ##f(z)## can be obtained from its characteristic function, ##g_z(t)## as follows:
$$
M_r = (-i)^r \left[ \frac{d^r g_z(t)}{ d t^r} \right]_{t=0}
$$
However, I haven't been able to obtain the result for ##M_1## myself because I have not been able to get a formula for the derivative of ##G_{\bar{y}}(t)##.

My approach to finding the derivative​

The first thing that struck me about the formula Schwartz gives for ##M_1## is that the summation is shown as being from ##k=1##, rather than from the ##k=-\infty## that appears in the formula for ##G_{\bar{y}}(t)##. That makes me think that there has been some kind of simplification based on symmetry, but I'm not sure what ... or how it has been used.

The second observation is that the formula inside the summation can be treated as the product of a ##\mathrm{sinc}## function and an exponential. I can obtain the derivative with respect to $t$ of both those parts, and use the chain rule to obtain the derivative of the product, which I have done ... and verified the result using Mathematica.

However, I'm not sure how then to proceed. Specifically, I don't know how deal with the summation over the product, and the subsequent raising to the power ##n##. Nor do I see what I would then do to progress to a formula with the changed lower bound on the summation.

You can almost always rewrite [tex]\sum_{k=-\infty}^{\infty} a_k = a_0 + \sum_{k=1}^\infty (a_k + a_{-k}).[/tex]

To calculate the derivative, I would abstract the details until they are needed. So start with [tex]
F(t) = \left(\sum_{k=-\infty}^{\infty} f_k(t)\right)^n[/tex] and proceed to [tex]
F'(t) = n\left(\sum_{k=-\infty}^\infty f_k(t)\right)^{n-1} \sum_{k=-\infty}^\infty f_k'(t)
[/tex] and hence [tex]\begin{split}
F'(0) &= n\left(\sum_{k=-\infty}^\infty f_k(0)\right)^{n-1} \sum_{k=-\infty}^\infty f_k'(0) \\
&= nF(0)^{(n-1)/n} \left( f_0'(0) + \sum_{k=1}^\infty (f_k'(0) + f_{-k}'(0))\right).
\end{split}
[/tex] Now you see that you need only caclulate [itex]f'_0(0)[/itex] and the combination [itex]f'_k(0) + f'_{-k}(0)[/itex].

Thank you very much for the clarity of your suggestions. I'll see how I go working the problem through , and let you know.

Now that I've been working on it, I see that there is something I didn't notice at first; something I realise I don't understand.
Please could you explain why the exponent of the ##F(0)## term in the second line of the identity
\begin{split}

F'(0) &= n\left(\sum_{k=-\infty}^\infty f_k(0)\right)^{n-1} \sum_{k=-\infty}^\infty f_k'(0) \\

&= nF(0)^{(n-1)/n} \left( f_0'(0) + \sum_{k=1}^\infty (f_k'(0) + f_{-k}'(0))\right),

\end{split}
is now ##(n-1)/n##, and not ##(n-1)##.

First steps, which I hope I have right. Evaluate ##F(0)## ...

Recognize that:

• ##f_k## involves the product of a ##\mathrm{sinc}## function and an exponential.
• Once we set ##t=0##, the ##\mathrm{sinc}## function reduces to ##\mathrm{sinc}(\pi k)##
• The summation variable ##k## is always an integer, and since ##\mathrm{sinc}(\pi k)## is equal to zero for all values of ##k## other than ##k=0## (at which point ##\mathrm{sinc}(0) = 1##, we also only need to consider ##k=0## in the exponential term.
• In the exponential term, when ##k=0##, the whole exponent is equal to zero, and the term is equal to 1.

Hence ##F(0)=1##.

My main uncertainty here is with regard to the ##\mathrm{sinc}## function. Is it reasonable to treat the value at 0 as being defined? I tend to think of ##\mathrm{sinc(x)}## as having a limiting value of 1 as ##x\rightarrow 0## but I'm a bit unsure about treating the value as actually being 1 in this context.

Next step, tackling the derivative ##f_{k}'## ...

pasmith said:

You can almost always rewrite [tex]\sum_{k=-\infty}^{\infty} a_k = a_0 + \sum_{k=1}^\infty (a_k + a_{-k}).[/tex]

To calculate the derivative, I would abstract the details until they are needed. So start with [tex]
F(t) = \left(\sum_{k=-\infty}^{\infty} f_k(t)\right)^n[/tex] and proceed to [tex]
F'(t) = n\left(\sum_{k=-\infty}^\infty f_k(t)\right)^{n-1} \sum_{k=-\infty}^\infty f_k'(t)
[/tex] and hence [tex]\begin{split}
F'(0) &= n\left(\sum_{k=-\infty}^\infty f_k(0)\right)^{n-1} \sum_{k=-\infty}^\infty f_k'(0) \\
&= nF(0)^{(n-1)/n} \left( f_0'(0) + \sum_{k=1}^\infty (f_k'(0) + f_{-k}'(0))\right).
\end{split}
[/tex] Now you see that you need only caclulate [itex]f'_0(0)[/itex] and the combination [itex]f'_k(0) + f'_{-k}(0)[/itex].

Thank you so much! You'll see I did post a couple of responses earlier but I now realize I was replying to myself instead of to you). When I first read your guidance I thought I would inevitably have to come back to ask for more help. But in fact, with very slow and careful work following your approach, I have been able to derive the formula for the first moment. Quite a thrill!

The trickiest part was when I was confronted with an exponential that doesn't appear in Schwartz's paper, but I then remembered my exponential/trigonometric identities and change the pesky exponential to a term in ##cos()+i\,sin()##. The final step was working out exactly how the different pieces combined as they were summed over the negative and positive and ranges of k (i.e. non-zero). The ##cos()## term disappeared, leaving a term in ##i\,sin()##, and the imaginary number itself disappears when the required product for the first-moment is calculated.

I'm now going to tackle the second moment ... for which I might have to look for some more help.