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Let $X\sim \mu$ be a random variable taking value on $R$. Assume that $d\mu=C\sqrt{4-x^2}dx $ for some normalized constant $C>0$. Assume that the limit of moment generating function $$ \lim_{t\to \infty}\frac{E[e^{tX}]}{t^{-3/2}e^{2t}}=1. $$

Can we use the above result to get the asymptotic limit of the derivative of the moment generating function $\frac{d}{dt} E[e^{tX}]=E[Xe^{tX}]$?

I try to find a relation between $E[Xe^{tX}]$ and $E[e^{tX}]$.

Note that $$ E[e^{tX}]=C\int_{-2}^2e^{tx}\sqrt{4-x^2}dx. $$

Hermi
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    In general, one cannot differentiate asymptotic relations, but you may want to have a look at Tauberian theorems. – nejimban Sep 12 '22 at 03:53
  • @nejimban Yes, I know. Is there any other method? – Hermi Sep 12 '22 at 03:54
  • I just saw your edits. You actually know that $X$ has a semicircular distribution. The Laplace transform (moment generating function) is given by a Bessel function. It is a special function but derivatives and asymptotics of Bessel functions are known. – nejimban Sep 12 '22 at 04:11
  • @nejimban Thanks. But how to get this one is given by a Bessel function? Is there any reference? – Hermi Sep 12 '22 at 15:41
  • This paragraph on Wikipédia gives the characteristic function ($R=2$ in your case) and cites the sources [1], [2], [3], and §9.6.18. I guess one way to show that $E[\mathrm e^{tX}]=\frac{J_1(2t)}t$ is to check that both $E[\mathrm e^{tX}]$ and $\frac{J_1(2t)}t$ (as functions of $t$) satisfy the same characteristic differential equation (with appropriate initial value/boundary conditions). – nejimban Sep 12 '22 at 17:42
  • FYI: $$\lim_{t\to\infty}\frac{E[X\mathrm e^{tX}]}{t^{-3/2},\mathrm e^{2t}}=\frac1{\sqrt\pi},$$ so probably Tauberian theorems cannot apply here (at least it does not match the derivative of the asymptotics of $E[\mathrm e^{tX}]$)… – nejimban Sep 12 '22 at 17:47
  • @nejimban Thank you very much! I think I can get the limit of $E[e^{tX}]$ by $J_1(2t)/t$ now. But how about the $\frac{d}{dt} E[e^{tX}]=-t^{-2}J_1(2t)+2t^{-1}J_1'(2t)$? Here we need the asymptotic limit of $J_1'(2t)$. I only found the asymptotic limit of $J_1(2t)$ in https://math.stackexchange.com/questions/177118/about-the-asymptotic-formula-of-bessel-function – Hermi Sep 12 '22 at 17:47
  • Good. I think $\frac{\mathrm d}{\mathrm dt},E[\mathrm e^{tX}]$ can be simplified to $$\frac{2 I_2(2 t)}{t^2}+\frac{4 I_3(2 t)}{t}$$ using the differential equation satisfied by the Bessel function $J_1$. – nejimban Sep 12 '22 at 17:50
  • (Sorry I mixed up characteristic function and m.g.f in an earlier comment. I hope you can wrap your head around this, I am not fan of tedious computations :).) – nejimban Sep 12 '22 at 17:54
  • @nejimban Sorry, I did not get your $d E[e^{tX}]/dt$. I found $J_0(x)-J_3(x)=2J_1'(x)$. – Hermi Sep 12 '22 at 18:01
  • Also, I found $I_1'(x)=(I_0(x)+I_3(x))/2$, then $\frac{d}{dt}E[e^{tX}]=I_0(2t)/t-I_1(2t)/t^2+I_3(2t)/t$.. – Hermi Sep 12 '22 at 18:06
  • @nejimban So it seems that $I_2(x)\sim x^{-1/2}e^x$ and $I_3(x)\sim x^{-1/2}e^x$ as $x\to \infty$ in https://math.stackexchange.com/questions/2791588/leading-terms-in-asymptotic-expansion-of-modified-bessel-function-of-the-first-k? – Hermi Sep 12 '22 at 18:12

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After giving it a try I found that it's probably easier to deal with hypergeometric functions (which encompass all Bessel functions): start by representing $X$ as $2(2Y-1)$ where $Y$ has a Beta$(\frac32,\frac32)$ distribution. Then $$M(t):=E\!\left[\mathrm e^{tX}\right]=\mathrm e^{-2t}E\!\left[\mathrm e^{4tY}\right]=\mathrm e^{-2t}\cdot{}_1F_1(\tfrac32;3;4t),$$ where $_1F_1$ is Kummer's confluent hypergeometric function. Its asymptotic behavior is known: $$_1F_1(\tfrac32;3;4t)\sim\frac{\Gamma(3)}{\Gamma(\tfrac32)}\cdot\mathrm e^{4t}(4t)^{-\frac32}=\frac{\mathrm e^{4t}t^{-\frac32}}{2\sqrt\pi}.\tag{1}$$ Therefore $$\lim_{t\to\infty}\frac{M(t)}{t^{-\frac32}\mathrm e^{2t}}=\color{red}{\frac1{2\sqrt\pi}}$$ (and not $1$). For the derivative, we have $$M'(t)=\mathrm e^{-2t}\left(-2\cdot{}_1F_1(\tfrac32;3;4t)+4\cdot\partial_t\,{}_1F_1(\tfrac32;3;4t)\right).$$ The derivative of Kummer's function gives $$ \partial_t\,{}_1F_1(\tfrac32;3;4t)=-\frac12\cdot{}_1F_1(\tfrac32;4;4t)+{}_1F_1(\tfrac32;3;4t), $$ where ${}_1F_1(\tfrac32;4;4t)=O\bigl(\mathrm e^{4t}t^{-\frac52}\bigr)$ is negligible as $t\to\infty$ compared to (1). Then $$ \partial_t\,{}_1F_1(\tfrac32;3;4t)\sim{}_1F_1(\tfrac32;3;4t).$$ Thus we can conclude that $M'(t)\sim 2M(t)$, and $$\lim_{t\to\infty}\frac{E[X\mathrm e^{tX}]}{t^{-3/2}\,\mathrm e^{2t}}=\frac1{\sqrt\pi}.$$

nejimban
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