# Derivative problem.

• Jul 12th 2009, 05:39 AM
Derivative problem.
I have a standard normal distribution. cdf stands for cumulative distr. function and pdf for probability distribution function.

since the cdf is an intergral of pdf, the pdf is the derivative of cdf.

I have the function of the form:

f(x) = cdf(x)-cdf(-x)

I want a 1st derivative of this function, but if I just use
f'(x) = pdf(x) - pdf(-x) it is always zero, since the distribution is symmetrical.

This makes no sense to me, the derivative should be zero for x where pdf(x)~0, but not zero near zero.

Where's my mistake?

• Jul 12th 2009, 06:47 AM
matheagle
The cdf is $F_X(x)=P(X\le x)$ is any situation.

So cdf(x)-cdf(-x) equals $P(X\le x)-P(X\le -x)=P(-x where I hope x>0.

In the continuous case this is $\int_{-x}^x f_X(t)dt$ where f(t) is the density.

By calc one the derivative is $f_X(x)+f_X(-x)$
• Jul 12th 2009, 06:58 AM
CaptainBlack
Quote:

I have a standard normal distribution. cdf stands for cumulative distr. function and pdf for probability distribution function.

since the cdf is an intergral of pdf, the pdf is the derivative of cdf.

I have the function of the form:

f(x) = cdf(x)-cdf(-x)

I want a 1st derivative of this function, but if I just use
f'(x) = pdf(x) - pdf(-x) it is always zero, since the distribution is symmetrical..

$f'(x)=\frac{d}{dx} cdf(x) - \frac{d}{dx} cdf(-x)=$ $\frac{d}{dx} cdf(x) - \frac{d}{dx}(-x) \left. \frac{d}{du} cdf(u)\right|_{u=-x}=pdf(x)+pdf(-x)=2\; pdf(x)$

Which we could have got to much faster by writing:

$f(x)=\int_{-x}^x pdf(u)\ du=2 \int_0^x pdf(u)\ du$

Then the fundamental theorem of calculus gives:

$f'(x)=2\; pdf(x)$

CB