Find argmax $\frac{1}{2b}(\frac{x}{Na})^{\frac{N-1}{2}}e^{-\frac{1}{2b}(x+Na)}\frac{1}{\sqrt{2\pi(N-1)}}(\frac{\frac{e}{b}\sqrt{Nax}}{2(N-1)})^{N-1}$ 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 $147.496$.
