Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a random variable $X$ if we have a pdf $f(x)$, then Continuous Random Variable is

$$F(x) = \int_{-\infty}^{x}f(t)dt$$ Next $F'(x) = \frac{d}{dx}F(x)= f(x)$

I don't follow this, $$\frac{d}{dx}F(x) = \frac{d}{dx} \int_{-\infty}^{x}f(t)dt = \int_{-\infty}^{x} \frac{\partial}{\partial x} f(t)dt $$

I don't see how that last step yields $f(x)$

share|cite|improve this question
you differentiate w.r.t. the upper limit as it is not an integral with a parameter. Also, the passage "... then Continuous Random Variable is" is remarkable by itself. Is it from the book? – Ilya Nov 30 '12 at 20:42
Oh okay woops thank you – Hawk Nov 30 '12 at 20:43
up vote 4 down vote accepted

This is just the Fundamental Theorem of Calculus.

A PDF (of a univariate distribution) is a function defined such that it is 1.) everywhere non-negative and 2.) integrates to 1 over $\Bbb R$.

If we define $F(x) = \int_{-\infty}^x f(t)\ dt$, then the Fundamental Theorem of Calculus gives you the desired result.

This function, $F(x)$, is called the "cumulative distribution function," or CDF. It is defined in this manner, so the relationship between CDF and PDF is not coincidental -- it is by design.

Note that your last step is incorrect -- $x$ is the independent variable of the derivative there, and it is also the upper limit of the integral (so the resulting integral will be a function in terms of $x$). You can't move the $d/dx$ inside the integral.

share|cite|improve this answer

You can see this by differentiating under the integral sign, which follows from the fundamental theorem of calculus:

$$ \frac{d}{dx} F(x) =\lim_{c\to-\infty} \frac{d}{dx} \int^{x}_{c} f(t) dt = f(x).1 -\lim_{c\to-\infty} f(c).\frac{dc}{dx} + \lim_{c\to-\infty}\int^{x}_{c} \frac{d}{dx} f(t) dt $$

Since $c\to-\infty$ is a constant, the second term disappears, and since $f$ is a function of $t$, $\frac{d}{dx} f(t)$ also disappears.

share|cite|improve this answer
I've never seen this kind of expansion before. Would you care justifying it a bit? – user88595 Apr 16 '14 at 17:22 read the 1st part of proof, then you can understand but you are assumed to know mean value theorem and squezee theorem. hope it helps.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.