1. ## 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?

2. The cdf is $\displaystyle F_X(x)=P(X\le x)$ is any situation.

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

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

By calc one the derivative is $\displaystyle f_X(x)+f_X(-x)$

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..
$\displaystyle f'(x)=\frac{d}{dx} cdf(x) - \frac{d}{dx} cdf(-x)=$$\displaystyle \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:

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

Then the fundamental theorem of calculus gives:

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

CB