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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 $f(x)$

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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 –  jip Nov 30 '12 at 20:43

2 Answers 2

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.

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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.

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I've never seen this kind of expansion before. Would you care justifying it a bit? –  user88595 Apr 16 at 17:22

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