0
$\begingroup$

I would like to prove the following inequality involving the parabolic cylinder function $D_{-\nu}(x)$ for $\nu>0$:

$$ \frac{2\hspace{1pt}D_{-\nu+1}(x)}{D_{-\nu}(x)} \;<\; x+\sqrt{x^2+4\hspace{1pt}(\nu+1)} ,\quad x\ge 0 $$

The inequality seems to be tight as $x\to\infty$ (for fixed $\nu$), and as $\nu\to\infty$ (for fixed $x$).

I managed to reduce the proof of monotone likelihood ratio of the fixed input correlation coefficient distribution to this inequality.

Edit: I found a paper that actually proves this: https://onlinelibrary.wiley.com/doi/epdf/10.1111/sapm.12401.

$\endgroup$
4
  • $\begingroup$ fx = 2*ParabolicCylinderD[-v + 1, x]/ParabolicCylinderD[-v, x] - x - Sqrt[x^2 + 4*(v + 1)]; fx /. {v -> 1} // Plot[#, {x, 38, 40}, PlotRange -> Full] & fx /. {v -> 1, x -> 38.477} I found that the inequality doesn't hold for v -> 1, x -> 38.477, did I make some mistakes? $\endgroup$
    – 138 Aspen
    Commented Jul 17 at 1:47
  • $\begingroup$ @138Aspen Looking at the plot, it becomes numerically unstable around x=38. So it's just a numerical issue with Mathmatica implementation default precision for large x. $\endgroup$
    – japalmer
    Commented Jul 18 at 2:10
  • $\begingroup$ fx = 2*ParabolicCylinderD[-v + 1, x]/ParabolicCylinderD[-v, x] - x - Sqrt[x^2 + 4*(v + 1)] // N; fx /. {v -> 1} // Plot[#, {x, 50, 60}, PlotRange -> Full] & fx /. {v -> 1, x -> 54.41} // N what about v -> 1, x -> 54.41? $\endgroup$
    – 138 Aspen
    Commented Jul 18 at 2:40
  • $\begingroup$ 54 > 38. I'm saying it becomes unstable for large x, starting around x=38. I can't add the plot in comment, but if you plot fx, you will see it start to oscillate randomly around x=38. $\endgroup$
    – japalmer
    Commented Jul 18 at 2:52

0

You must log in to answer this question.