3
$\begingroup$

I am interested in modifying the Laplace distribution to take the general form $\textrm{poly}(|x|)\exp(-|x|)$, where $\textrm{poly}(\cdot)$ denotes a function of finite polynomial degree. In particular, I am interested in the class of densities which can be written as

$$ \frac{1}{2(\alpha+1)}\sum^{\alpha}_{k=0} \frac{|x|^k}{k!}e^{-|x|}, $$

where integer $\alpha \ge 1$.

Interestingly, the derivative has the simple form

$$ \frac{sign(x)}{2(\alpha+1)}\Big(e^{-|x|}\sum^{\alpha}_{k=0}\frac{k|x|^{k-1}}{k!}-e^{-|x|}\sum^{\alpha}_{k=0}\frac{|x|^k}{k!}\Big) = -\frac{e^{-|x|}}{2(\alpha+1)!}x|x|^{\alpha-1}. $$

However, I am struggling to find the integral for this class of densities. It's straightforward enough to put a particular density, say for $\alpha=2$, into WolframAlpha, but I don't know how to obtain the general form for any $\alpha$.

$\endgroup$
1
  • 3
    $\begingroup$ The absolute value $|x|$ makes it hard to integrate, but $\int x^ke^{-x} = -\Gamma(k+1,x)$. $\endgroup$
    – Ricky
    Aug 28, 2023 at 3:43

1 Answer 1

2
$\begingroup$

If $P$ is a polynomial and let for all $x \in \mathbb{R}$, $$ f(x) = \int_0^x P(|t|)e^{-|t|}\, dt, \qquad g(x) = \int_0^x P(t)e^{-t} \, dt. $$ Then, by parity, we clearly have $f(x) = \mathrm{sgn}(x)g(|x|)$ for all $x$ and if $t \mapsto P(|t|)e^{-|t|}$ is the distribution of a probability distribution $\mu$, then for all $a < b$, $\mu([a,b]) = f(b) - f(a) = \mathrm{sgn}(b)g(|b|) - \mathrm{sgn}(a)g(|a|)$. Therefore, it is enough to study $g$ on $\mathbb{R}_+$.

A first method to compute $g$, if you know the coefficients of $P$ (like in your case), is to use the incomplete Gamma function, as @Ricky suggested it. An other one is to write $P$ in an other base of $\mathbb{R}[x]$. Indeed, for all $k$, $\partial_x(x^ke^{-x}) = kx^{k - 1}e^{-x} - x^ke^{-x}$ so if you set $b_k(x) = -x^k + kx^{k - 1}$ (so $b_0 = -1$), you have, $$ \int_0^x b_k(t)e^{-t} \, dt = [t^ke^{-t}]_0^x = x^ke^{-x} - \delta_{0k}. $$ The $b_k$ clearly form a base of $\mathbb{R}[x]$ (and $(b_0,\ldots,b_n)$ is a basis of $\mathbb{R}_n[x]$) so you can write $P(x) = \sum_{k = 0}^n a_kb_k(x)$ where $n = \deg(P)$ and you have, $$ g(x) = a_0e^{-x} - a_0 + \sum_{k = 1}^n a_kx^ke^{-x}. $$ In your case, you can use the simple form of the derivative of $P$ to get that, \begin{align*} g(x) & = \int_0^x P(t)e^{-t} \, dt\\ & = [tP(t)e^{-t}]_0^x - \int_0^x t\partial_t(P(t)e^{-t}) \, dt \textrm{ by IBP,}\\ & = xP(x)e^{-x} + \frac{1}{2(\alpha + 1)!}\int_0^x t^{\alpha + 1}e^{-t} \, dt\\ & = \frac{1}{2(\alpha + 1)}\sum_{k = 0}^\alpha \frac{x^{k + 1}}{k!}e^{-x} + \frac{\gamma(\alpha + 2,x)}{2(\alpha + 1)!}. \end{align*} Therefore, if $\mu$ is the associated measure, for all $a < b$, $$ \mu([a,b]) = \frac{\mathrm{sgn}(b)}{2(\alpha + 1)}\sum_{k = 0}^\alpha \frac{|b|^{k + 1}}{k!}e^{-|b|} + \frac{\mathrm{sgn}(b)\gamma(\alpha + 2,|b|)}{2(\alpha + 1)!} - \frac{\mathrm{sgn}(a)}{2(\alpha + 1)}\sum_{k = 0}^\alpha \frac{|a|^{k + 1}}{k!}e^{-|a|} - \frac{\mathrm{sgn}(a)\gamma(\alpha + 2,|a|)}{2(\alpha + 1)!} $$

$\endgroup$
2
  • $\begingroup$ Just to make sure I understand correctly, the CDF at $t$ is then $g(|t|)-g(|-\infty|) = \lim_{a \rightarrow -\infty} \mu([a, b])$? $\endgroup$
    – calmcc
    Aug 28, 2023 at 12:48
  • 1
    $\begingroup$ I made a sign error that I corrected in the expression of $f$ so $\mu(]-\infty,b]) = f(b) - \lim_{-\infty} f = \mathrm{sgn}(b)g(|b|) + \lim_{+\infty} g$. $\endgroup$
    – Cactus
    Aug 29, 2023 at 8:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .