I would like to ask you for a help with my problem. I have random variable $$X = X_1 + \sum_{i = 2}^{n}(X_i - y_i)^{+}, \ \ \ (.)^{+} = \max(0, . ), $$ where $X_i$ are independent, continuous with known densities, CDFs. And I would like to perform a derivative of $$F_X(g(y_2))$$ w.r.t. $y_2$ and $g(x)$ is well defined continuous smooth function.

My tries looked like \begin{equation} \begin{aligned} \frac{\partial}{\partial y_2}F_X(g(y_2)) &= \frac{\partial}{\partial y_2}\left[ F_{X | X_2 > y_2}(g(y_2)) \ \mathbb{P}(X_2 > y_2) + F_{X | X_2 \leq y_2}(g(y_2)) \ \mathbb{P}(X_2 \leq y_2)\right] =\\ &= f_{\bar{X} | X_2 > y_2}(g(y_2) + y_2) \ \mathbb{P}(X_2 > y_2)\left( \frac{\partial g(y_2)}{\partial y_2} + 1\right) - F_{X | X_2 > y_2}(g(y_2)) f_{X_2}(y_2) \\ &+ f_{\bar{X} | X_2 \leq y_2}(g(y_2)) \left( \frac{\partial g(y_2)}{\partial y_2} \right) \ \mathbb{P}(X_2 \leq y_2) + F_{X | X_2 \leq y_2}(g(y_2)) f_{X_2}(y_2), \end{aligned} \end{equation} where $\bar{X}$ is adjusted r.v. $X$ such that I move $y_2$ on RHS.

Please could anyone tell me if my steps are correct. I am not sure if I can derive the conditional CDF in the "standard" way because the condition $X_2 > y_2$ still depends on $y_2$. I will be glad for any idea.


closed as off-topic by gung May 1 '18 at 14:12

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  • $\begingroup$ puzzlement: I do not understand the limitation to $X_2$ only as the $y_2$ used in the argument of the cdf is not connected with the $y_2$ used in the definition of $X$. $\endgroup$ – Xi'an May 1 '18 at 10:16
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    $\begingroup$ Please do not cross-post:math.stackexchange.com/questions/2760950/…. $\endgroup$ – StubbornAtom May 1 '18 at 10:39
  • $\begingroup$ @Xi'an the $y_2$ are the same in both cases. This question arises as the fact that I am looking for $y^* = g(y_2,y_3,...)$ which fulfills $F_X(y^*) = 0$ and I am wondering how does $y^*$ changes as $y_2$ changes. So I am trying to employ implicit derivation in order to get the desired relation. $\endgroup$ – Ajvn May 1 '18 at 12:11
  • $\begingroup$ I'm voting to close this question as off-topic because it is cross posted on Mathematics. $\endgroup$ – gung May 1 '18 at 14:12
  • $\begingroup$ I still do not understand the validation of this differentiation. $\endgroup$ – Xi'an May 1 '18 at 20:02