Timeline for Taylor series expansion of quantile function
Current License: CC BY-SA 4.0
12 events
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| Feb 21, 2019 at 14:24 | comment | added | Synia | You can have a look at en.wikipedia.org/wiki/Lagrange_inversion_theorem for that. Nevertheless, your problem seems more subtle if you do not assume a nice expansion in $ \lambda $. I would also be interested in an expression of the coefficients $ \frac{d^n g}{d\lambda^n}(0) $ in terms of, say, moments. This looks like an interesting way of estimating a quantile. | |
| Feb 21, 2019 at 8:29 | comment | added | NN2 | Ah I see. The problem with this view become easy and evident, I don't know why I made it complicated. Thank you very much for your answer Robert! | |
| Feb 20, 2019 at 19:03 | comment | added | Robert Israel | Moreover, you can find the coefficients of the series of $g(\lambda)$ from the series for $F_{X+\lambda Z}(\alpha)$ by reversion of series. | |
| Feb 20, 2019 at 18:41 | comment | added | Robert Israel | $F^{-1}_{X+\lambda Z}(\alpha)$, for fixed $\alpha$ is some function of $\lambda$: call it $g(\lambda)$ for short. Then, if this is sufficiently differentiable, Taylor says $g(\lambda) = \sum_{n=0}^N \frac{d^n g}{d\lambda^n}(0) \frac{\lambda^n}{n!} + O(\lambda^{N+1})$. | |
| Feb 20, 2019 at 16:57 | comment | added | NN2 | Thanks Matt F, I suppose $Y$ and $Z$ follow normal distribution and can confirm that this formula is correct. In fact, $F^{-1}_{Y+\lambda Z}(\alpha) = q_{\alpha}(\sigma_X^2 +\lambda^2\sigma_Y^2)^\frac{1}{2}$. But how to prove this formula in general case, when we don't have explicite formula of quantile function? I have no idea. | |
| Feb 20, 2019 at 16:12 | comment | added | user44143 | An example may help: Suppose X and Y are both lognormal, eg modeling returns for bonds and stocks, so lambda measures the riskiness of the portfolio. What are good series for approximating the quantile function near a particular portfolio value? | |
| Feb 20, 2019 at 15:36 | comment | added | NN2 | The approximation of a usual Taylor formula is usually at the value of $\alpha$ and not at the functions. So, this formula is not evident for me and I don't know how to prove it (or find the coefficients). Yes, we can suppose that $Y$ and $Z$ have a joint distribution with a density. | |
| Feb 20, 2019 at 15:29 | comment | added | Robert Israel | To have any hope of analyticity, I think you'll want to assume $Y$ and $Z$ have a joint distribution with a density. | |
| Feb 20, 2019 at 15:21 | comment | added | Robert Israel | This is just the usual formula for a Taylor series. Or do you mean you want to find the coefficients? | |
| Feb 20, 2019 at 15:20 | history | edited | András Bátkai | CC BY-SA 4.0 |
added arxiv tag
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| Feb 20, 2019 at 15:15 | review | First posts | |||
| Feb 20, 2019 at 15:20 | |||||
| Feb 20, 2019 at 15:14 | history | asked | NN2 | CC BY-SA 4.0 |