The following problem is what I am working on.

$F(x)=\frac{1}{1+e^{-x}}$ is the cumulative density function defined for all real numbers. Find the probability density function.

My understanding is that $\int_{-\infty}^{x} f(t)dt=F(x)$. So I understand that taking the derivative with respect to $x$ on both sides is what I would like to do, but what do we do with the $-\infty$ ? The book that I am working on explains it as just take the derivative of $F(x)$, but as a math instructor I am hesitant to blindly follow that.

Can someone help ?


Differentiating $\int_{-\infty}^x f(t) dt$ and $\int_{a}^x f(t)dt$ for any real $a$ will give you the same function $f(t),$ since these two functions differ by a constant value $\int_{-\infty}^a f(t)dt$ only. That $d/dx(\int_a^x f(t)dt)= f(x)$ is the content of the fundamental theorem of calculus.

  • $\begingroup$ Thank you very much ! I should have brute forced my way to get to your solution. It makes so much sense ! $\endgroup$ – hyg17 Jan 16 '14 at 20:00

We have that: $$ \dfrac{\partial }{\partial x} \int^{b(x)}_{a(x)} f(z)dz= b^\prime(x)\cdot f(x)-a^\prime(x)\cdot f(x).$$ In your example, $b(x)=x$ so $b^\prime(x)=1$ and $a(x)=-\infty$ is a constant so $a^\prime(x)=0$ then you get $$f(x)=F^\prime(x).$$

  • $\begingroup$ Your $a(x)$ is not a real valued function. The concern is that integrating from $-\infty$ to $x$ is not technically a Riemann integral to which the fundamental theorem of calculus directly applies, since there is a hidden limit. $\endgroup$ – B. Mackey Jan 16 '14 at 19:55
  • $\begingroup$ Thank you! I like your precise explanation ! $\endgroup$ – hyg17 Jan 16 '14 at 20:01

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