TITLE: Derivative of error function and relationship between cumulative distribution function and error function
QUESTION [0 upvotes]: to my understand for part a), the derivative is simply replacing $t$ with x by the first fundamental theorem of calculus then computing erf'(3) into $1/ sqrt(pi) * exp(-x^2)$
for part b) however, I am not sure if what I have is correct. Since the question states $x >= 0$ , the only possible relationship between CDF and erf(x) I can think of is $Φ(x) = 1/2 (1+ erf(x/sqrt(2))$. Am I on the right track?
edit: this is what i obtained - not sure if this is correct

REPLY [1 votes]: If $X$ is   random variable with distribution $\Phi$ then, for $x \geq 0$, $P\{X\leq x\}=P\{X\leq 0\}+P\{0<X \leq x\}=\frac 1 2 +\int_0^{x}\frac 1 {\sqrt {2\pi}} e^{-t^{2} /2} \, dt$. To see its connection with the the formula for erf note that $erf(x)=\frac 2 {\sqrt 2\pi}\int_0^{x} e^{-t^{2} /2} \, dt$. So $P\{X\leq x\}=\frac 1 2 [1+erf (x)]$.
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