Lower bounds on the modified Bessel function of the first kind

Question

It is known that the modified Bessel function $I_{n}(x)$ satisfies the lower bound

\begin{eqnarray*} I_{n}(x) > \frac{1}{\Gamma(n+1)} \left( \frac{x}{2} \right)^n \end{eqnarray*}

for $x > 0$, $n > -\frac{1}{2}$. This lower bound is pretty good when $x$ is small, but it loses tightness when $x$ is about the same size as $n$; the plot below shows this discrepancy in a logarithmic scale for $I_{20}(x)$ (the dashed curve is the lower bound):

A better lower bound for $I_n(x)$ likely has to incorporate the term $e^{x}$ that is found in its asymptotic expansion:

\begin{eqnarray*} I_{n}(x) \sim \frac{e^{x}}{\sqrt{2 \pi x}}\left( 1 - \frac{4n^2 -1}{8x} + \cdots \right) \end{eqnarray*}

I have not been able to find such a lower bound thus far. To paraphrase, the bound I am seeking about is good for $x$ in the order of $n$, and I hypothesize that such a bound must include the term $e^x$.

Answer 1

You can get this for $n=0$ as follows (I needed it in my research just now). The result is $I_0(x) \ge \frac{2\sqrt{2}}{3\pi}\frac{e^x}{\sqrt{x}}$, for all $x\ge 1/2$. Take the integral rep. https://dlmf.nist.gov/10.32#E2 whose integrand is the ratio of two positive functions. The dominant contribution is from a region of size $1/x$ approaching $t=1$, and we can ignore the rest of the domain, and use simple tangent line bounds for the two functions. Use a lower bound $e^{xt} \ge e^x(1-(1-t)x))$, and the upper bound $1-t^2 \le 2(1-t)$. Change variable to $u=1-t$ and shrink the domain to $u\in(0,1/x)$, which must lie in $(0,2)$, giving the restriction on $x$. The integrals become simple algebraic ones, and you should get the constant I did, unless I made a mistake. The constant is about 0.3001 which is pretty close to the asymptotic $1/\sqrt{2\pi} \approx 0.3989$.

For $n>0$ the same integral rep should work, but I suggest you leave $e^{xt}$ exact, giving something like a Gamma function bound. Let me know if it works out, or if you improve/fix the above...

Answer 2

Expanding on the answer by Alex, a pretty good lower bound for all integer $n \geq 0$ and $x > 1$ can be obtained by using the inequality [1, eq. 9]

$$ \frac{I_n(x)}{I_{n-1}(x)} \geq \frac{\sqrt{x^2 + n^2}-n}{x}, \quad n \geq 1. $$

Applying this recursively yields

$$ I_n(x) \geq x^{-n}I_0(x)\prod_{k=1}^n (\sqrt{x^2 + n^2} - n). $$

The product can be bounded by taking its logarithm to convert it into a sum, and bounding it by an integral. Exponentiating again and cleaning it all up (I won't reproduce it here for readability), you end up with

$$ I_n(x) \geq C \cdot \left(\frac{\sqrt{x^2 + n^2} - n}{x}\right)^{n+1} \frac{e^{\sqrt{x^2 + n^2}}}{\sqrt{x}}, $$

for some constant $C > 0$. In the end there I just used Alex's bound for $x > 1$. It's also straightforward to extend this to fractional $n \geq 0$; use the process outlined by Alex in their comment for some $n' \in (0, 1)$, and replace $I_0$ by $I_{n'}$ in the above.

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[1] D. E. Amos, Computation of modified Bessel functions and their ratios, Mathematics of Computation 28 (1974), no. 125, 239–251