I want to find an expression for:
$$ (1) \quad \underset{x}{argmax} \Big \{ \frac{1}{2b} \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\Big( \frac{\sqrt{Nax}}{b} \Big) \Big \} $$
Where $a,b,x \in \mathbb{R_{++}}$, $N \in \mathbb{N}$, and $I_{N-1}(\cdot)$ is the modified Bessel function of the first kind of order $N-1$.
Since this function is concave in $x$, all I need to do is take the partial derivative w.r.t $x$, set it equal to zero, and solve for $x$. Unfortunately, this seems like a pretty intractable problem. Try computing the $x$ partial in Wolfram to see what I mean.
So, I am going to approximate the answer instead. I know from NIST that for large orders:
$$I_{N-1}(z) \sim \frac{1}{\sqrt{2 \pi (N-1)}} \Big( \frac{e z}{2 (N-1)} \Big)^{N-1}$$
So then I have:
$$ (2) \quad \underset{x}{argmax} \Big \{ \frac{1}{2b} \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} \frac{1}{\sqrt{2 \pi (N-1)}} \Big( \frac{\frac{e}{b} \sqrt{Nax} }{2 (N-1)} \Big)^{N-1} \Big \} $$
This problem is easy. After taking the logarithm, expanding terms, taking the derivative, and setting it to zero, I get:
$$ x^{\ast} = 2b (N-1) $$
This is exact for (2), but nowhere near the truth for (1), especially as $N$ gets large and $a,b \approx 1$. This can be checked numerically. I think this is because the NIST expansion doesn't account for the argument growing as well.
Questions:
- How can I solve (1) analytically? Can you solve it exactly via the derivative?
- Is there an expansion, similar to the one used in (2), for when both the order and argument of the Bessel function are growing?
- Any other ideas on how I can approach this problem? Maybe approximate the log of the Bessel function directly?
Thanks!
We simplify slightly the expression by writing \begin{align} &\frac{1}{2b} \left(\frac{x}{Na}\right)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)=\frac{1}{2b}\left( Na \right)^{\frac{1-N}{2}}e^{-\frac{Na}{2b}}f(x)\\ &f(x)=x^{\frac{N-1}{2}}e^{-\frac{x}{2b}} I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right) \end{align} Then, we want to obtain the solutions of $f'(x)=0$. The derivative reads \begin{equation} f'(x)=\frac{N-1}{2}\frac{f(x)}{x}-\frac{1}{2b}f(x)+\frac{\sqrt{Na}}{2b\sqrt{x}}\frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}f(x) \end{equation} The equation to solve is thus \begin{equation} \frac{N-1}{x}-\frac{1}{b}+\frac{\sqrt{Na}}{b\sqrt{x}}\frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}=0 \end{equation} When both the order and the argument are large, a uniform asymptotic expansion can be used for $I_{N-1}$ and $I_{N-1}$: \begin{align} I_{N-1}\left((N-1) z\right)&\sim\frac{e^{(N-1)\eta}}{(2\pi(N-1))^{\frac{1}{2}}(1+z^{2% })^{\frac{1}{4}}}\sum_{k=0}^{\infty}\frac{U_{k}(p)}{(N-1)^{k}}\\ I_{N-1}'\left((N-1) z\right)&\sim\frac{(1+z^{2})^{\frac{1}{4}}e^{(N-1)\eta}}{(2\pi% (N-1))^{\frac{1}{2}}z}\sum_{k=0}^{\infty}\frac{V_{k}(p)}{(N-1)^{k}} \end{align} where \begin{equation} \eta=(1+z^{2})^{\frac{1}{2}}+\ln\frac{z}{1+(1+z^{2})^{\frac{1}{2}}} \end{equation} Here, we choose \begin{equation} z=\frac{\sqrt{Nax}}{b(N-1)} \end{equation} Keeping the term $k=0$ only, we find (with $U_0(p)=V_0(p)=1$) \begin{equation} \frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}=\frac{\sqrt{1+z^2}}{z} \end{equation} The equation to be solved becomes \begin{equation} \frac{N-1}{x}-\frac{1}{b}+\frac{\sqrt{Na}}{b\sqrt{x}} \sqrt{1+\frac{b^2(N-1)^2}{Nax}}=0 \end{equation} or \begin{equation} N-1-\frac{x}{b}+ \sqrt{\frac{Nax}{b^2}+(N-1)^2}=0 \end{equation} which has a simple non-zero solution \begin{equation} x=Na+2(N-1)b \end{equation} which seems to be numerically correct. For $a=1,b=1,N=20$, the approximation gives $x=58$ while a numerical evaluation of the solution is $x=57.49$. For $a=1,b=1,N=50$ we find $148$ to be compared to $147.496$.