Derivative of cumulative function

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Discussion Overview

The discussion revolves around finding the derivative of a cumulative function defined as F(x) = P[Y ≤ g(x)], where Y is a random variable and g(x) is a function of x. The focus is on the mathematical formulation and differentiation of this probability function.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant asks for the formula to differentiate the cumulative function F(x) = P[Y ≤ g(x)].
  • Another participant suggests expressing F(x) in terms of the density function h(y) and the cumulative distribution function H(y), leading to F(x) = H(g(x)).
  • A participant proposes the derivative as F'(x) = dH[g(x)]/dg(x) · dg(x)/dx, noting that the sign of F'(x) depends on dg(x)/dx.
  • Another participant states that H'(g(x)) g'(x) can be expressed as h(g(x)) g'(x), indicating a relationship between the derivatives of the cumulative and density functions.

Areas of Agreement / Disagreement

Participants appear to agree on the formulation of the derivative but do not reach a consensus on the implications of the signs or the final expression for F'(x). The discussion remains somewhat unresolved regarding the clarity of the derivative's behavior.

Contextual Notes

There are assumptions regarding the properties of the functions involved, such as the positivity of the density function and the behavior of g(x). The discussion does not fully resolve the mathematical steps required for differentiation.

toltol
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Hi everybody,

Can someone tell me the formula to I should use to find the derivative of the following function, with respect to x:

F(x)=Probability[Y<=g(x)]

dF(x)/dx = ??

Thank you for your help.

Toltol
 
Last edited:
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Well, first step, let h(y) be the density function for Y, and let H(y) be the cumulative distribution function for Y. Now we have
F(x)=P(Y<=g(x))
=[tex]\int_{-\infty}^{g(x)} h(y) dy[/tex]
=H(g(x))
Now, can you differentiate that?
 
Thank you mXSCNT.

If F(x)=P[Y<=g(x)]=H[g(x)]

Thus, the derivative is:

F'(x)=dH[g(x)]/dg(x) . dg(x)/x

The term dH[g(x)]/dg(x) is >0; Thus, the sign of F'(x) is the sign of dg(x)/x.

Am I ok?

Thank you,
Toltol
 
H'(g(x)) g'(x) = h(g(x)) g'(x)
 
Thank you mXSCNT.

It's OK.
 

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