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Suppose I have a r.v. $Z = X + \alpha Y$ and that $F_Z$ is the probability distribution function of $Z$. If we think of the probability $p = F_Z(q) = \mathbb{P}(X+\alpha Y < q)$ as a function $p = p(q, \alpha)$, how can we write the derivative $\partial_{\alpha} p(q, \alpha)$ supposing as much regularilty on the distribution of $X$ and $Y$ as we want?

I'm loosely thinking of a situation where you know the marginals distributions $F_X$ and $F_Y$ and maybe locally (around $(q, \alpha)$) know some information about the dependence of $X$ and $Y$ (maybe correlation is enough in some cases to build an approximation?).

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  • $\begingroup$ en.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$
    – Douglas Zare
    Commented Jul 20, 2012 at 23:51
  • $\begingroup$ I guess I should say I'm looking for something where you don't use the full joint distribution function. $\endgroup$
    – safetyduck
    Commented Jul 22, 2012 at 1:24
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    $\begingroup$ Then you're looking for something that doesn't exist. $\endgroup$
    – Robert Israel
    Commented Jul 22, 2012 at 7:46
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    $\begingroup$ To see that correlation isn't enough, remember that adding outliers can affect the correlation coefficients with a negligible effect on probabilities like $P(X+\alpha Y \lt q)$. $\endgroup$
    – Douglas Zare
    Commented Jul 22, 2012 at 9:44

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If $X$ and $Y$ have joint density $f(x,y)$, we have $p(q,\alpha) = \int_{-\infty}^\infty dy\ \int_{-\infty}^{q - \alpha y} dx\ f(x,y)$ and thus (assuming sufficient regularity) $\partial_\alpha p(q,\alpha) = - \int_{-\infty}^\infty dy \ y f(q-\alpha y,y)$.

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  • $\begingroup$ The appendix here is useful: efinance.org.cn/cn/FEshuo/… $\endgroup$
    – safetyduck
    Commented Aug 7, 2012 at 18:56
  • $\begingroup$ ... and in case the internet changes by the time you read the previous comment the reference is: Sensitivity analysis of Values at Risk, C. Gourieroux, J.P. Laurent, O. Scaille $\endgroup$
    – safetyduck
    Commented Aug 7, 2012 at 19:43

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