Inverse Mellin transform of the Mellin transform of the binomial PMF

Question

The probability mass function of the Binomial distribution is given by

$$\begin{equation} f(x)=\binom{n}{x} p^x (1-p)^{n-x}, \end{equation}$$ where $p \in [0,1]$ and $x=\{0,1,\dots,n\}$ (finite support).

The Mellin transform of $f$ is therefore $$\begin{align} \mathcal{M}\{f\}(s)&=\int_0^{\infty}x^{s-1} \binom{n}{x} p^x (1-p)^{n-x}\mathrm{d}x\\ &=\sum_{k=0}^{n}k^{s-1} \binom{n}{k} p^k (1-p)^{n-k}. \end{align}$$ Since $f$ is a probability mass function, the integral becomes a sum. Now I'd like to apply the inverse Mellin transform to get back $f$: $$\begin{align} \mathcal{M}^{-1}\{\mathcal{M}\{f\}\}(x) &= \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} x^{-s} \left( \sum_{k=0}^{n}k^{s-1} \binom{n}{k} p^k (1-p)^{n-k} \right) \mathrm{d}s \\ &=\frac{1}{2 \pi i} \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \int_{c-i\infty}^{c+i\infty} x^{-s} k^{s-1} \mathrm{d}s. \end{align}$$

The problem is that the integral $\int_{c-i\infty}^{c+i\infty} x^{-s} k^{s-1} \mathrm{d}s$ doesn't seem to converge. Essentially it's equal to $$ \begin{align} \int_{c-i \infty}^{c+i\infty} x^{-s}k^{s-1} \mathrm{d} s &= \frac{1}{x} \int_{c-i \infty}^{c+i\infty} \left(\frac{k}{x}\right)^{s-1} \mathrm{d}s\\ &=\frac{i}{x} \int_{-\infty}^{\infty} e^{it \ln \left({\frac{k}{x}} \right)} \mathrm{d}t. \end{align} $$ So the integrand is forever oscillatory, and doesn't seem to have a value. I did read that it "equals" the dirac Delta distribution, but I'm not sure how it'll fit in with the other terms since that's not really a function. But at the same time, since $f$ has support on a finite set, it kind of makes sense for it to show up?

I'm pretty lost, how can I recover $f$?

Answer 1

Thanks to Steven Clark for some clarification. Indeed the dirac-Delta is very relevant here, I should've played around with it more.

What happens is that \begin{align} \int_{c-i \infty}^{c+i\infty} x^{-s}k^{s-1} \mathrm{d} s &= \frac{1}{x} \int_{c-i \infty}^{c+i\infty} \left(\frac{k}{x}\right)^{s-1} \mathrm{d}s,\\ &=\frac{i}{x} \int_{-\infty}^{\infty} e^{it \ln \left({\frac{k}{x}} \right)} \mathrm{d}t,\\ &= \frac{2 \pi i}{x} \delta\left(\ln\left(\frac{k}{x}\right)\right),\\ &= 2 \pi i \delta \left(k-x\right). \end{align}

and what I completely missed is that $\mathcal{M}^{-1}\{\mathcal{M}\{f\}\}(x)$ is a function of $x$ on the real number line, and that $\delta\left(\ln\left(\frac{k}{x}\right)\right) = x \delta \left(k-x\right)$, which not only cancels the $x$ in the denominator that I could not get rid of, but also means that the integral is 0 whenever $x \neq k$. So we get that \begin{align} \mathcal{M}^{-1}\{\mathcal{M}\{f\}\}(x) &=\frac{1}{2 \pi i} \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \int_{c-i\infty}^{c+i\infty} x^{-s} k^{s-1} \mathrm{d}s,\\ &=\frac{1}{x}\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \delta\left(\ln\left(\frac{k}{x}\right)\right),\\ &=\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \delta\left(k-x\right),\\ &=f(x). \end{align} Very cleverly, all the terms inside the sum equal zero except for when $x=k$, which is where the dirac-Delta spikes. This is exactly the behavior $f$ (and probability mass functions in general) would exhibit had it been defined on the entire real number line!

It also does integrate to 1 along $\mathbb{R}$, and the antiderative $\int_{-\infty}^x f(t) \mathrm{d}t$ is a step function that changes in value at each $x=\{0,1,\dots,n\}$ by $\binom{n}{x} p^x (1-p)^{n-x}$, which is indeed the binomial CDF.

Answer 2

Its not the Mellin transform of

$$f(x)=\binom{n}{x}\, p^x\, (1-p)^{n-x}\tag{1},$$

but rather the Mellin transform of

$$g(x)=\sum\limits_{k=0}^n f(x)\, \delta (x-k)\tag{2}$$

which, assuming the convention $\int\limits_0^\infty \delta(x)\, dx=1$ (versus $\int\limits_0^\infty \delta(x)\, dx=\frac{1}{2}$ for example), is

$$\mathcal{M}_x[g(x)](s)=\sum\limits_{k=0}^n \left(\int_0^{\infty} f(x)\, \delta (x-k)\, x^{s-1} \, dx\right)\\=\sum\limits_{k=0}^n f(k)\, k^{s-1}=\sum\limits_{k=0}^n \binom{n}{k}\, p^k\, (1-p)^{n-k}\, k^{s-1}\tag{3}$$

and this explains the presence of the Dirac delta terms in the inverse Mellin transform.

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But you still need to normalize it such that $$\sum\limits_{k=0}^n \left(\int\limits_{-\infty}^{\infty} f(x)\, \delta (x-k) \, dx\right)=\sum\limits_{k=0}^n f(k)=\sum\limits_{k=0}^n \binom{n}{k}\, p^k\, (1-p)^{n-k}=1\tag{4}.$$

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The remainder of this answer illustrates an alternate perspective on the problem.

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Assuming the definition

$$f(k)=\left\{\begin{array}{cc} \binom{n}{k}\, p^k\, (1-p)^{n-k} & 0\leq k\leq n \\ 0 & k>n \\ \end{array}\right.\tag{5}$$

consider the summatory step function

$$h(x)=\sum\limits_{k=0}^x f(k)\tag{6}$$

which can be viewed as sort of a Cumulative Distribution Function (CDF) with first-order derivative

$$h'(x)=\sum\limits_{k=0}^x f(k)\,\delta(x-k)\tag{7}$$

which can be viewed as sort of a Probability Density Function (PDF).

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Now consider the Mellin transform

$$H(s)=s\, \mathcal{M}_x[h(x)](-s)=s\, \int_0^{\infty} h(x)\, x^{-s-1} \, dx\\=2 \pi\, f(0)\, s\, \delta(-i s)+\sum\limits_{k=1}^n \frac{f(k)}{k^s}\tag{8}$$

which is equivalent to the Mellin transform

$$H(s)=\mathcal{M}_x[h'(x)](1-s)=\int_0^{\infty} h'(x)\, x^{-s} \, dx\\=2 \pi\, f(0)\, s\, \delta(-i s)+\sum\limits_{k=1}^n \frac{f(k)}{k^s}\tag{9}$$

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The function $h(x)$ is recovered from the inverse Mellin transform

$$h(x)=\mathcal{M}^{-1}_s\left[\frac{H(s)}{s}\right]\left(\frac{1}{x}\right)=\frac{1}{2 \pi i} \int_{\alpha-i \infty}^{\alpha+i \infty} \frac{H(s)}{s}\, x^s \, ds=\sum\limits_{k=0}^x f(k)\tag{10}$$

where $\alpha=0$ must be used for the $2 \pi\, f(0)\, s\, \delta(-i s)$ term of $H(s)$ and $\alpha>0$ must be used for the remaining terms of $H(s)$.

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Formulas (6) to (10) above are more consistent with the way a summatory step function

$$r(x)=\sum\limits_{n=1}^x a(k)\tag{11}$$

is related to its corresponding Dirichlet series

$$R(s)=\sum\limits_{n=1}^\infty \frac{a(k)}{k^s}\tag{12}.$$