这是一份kcl伦敦大学学院 4CCM112A作业代写的成功案
A famous distribution function is given by the standard normal distribution, whose probability density function $f$ is
$$
f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}, \quad \text { for }-\infty<x<\infty
$$
This is often called a “bell curve”, and is used widely in statistics. Since we are claiming that $f$ is a p.d.f., we should have
$$
\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x=1
$$
by formula , which is equivalent to
$$
\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x=\sqrt{2 \pi}
$$
4CCM112A COURSE NOTES ：
The expected value $E X$ of a random variable $X$ can be thought of as the “average” value of $X$ as it varies over its sample space. If $X$ is a discrete random variable, then
$$
E X=\sum_{x} x P(X=x) \text {, }
$$
with the sum being taken over all elements $x$ of the sample space. For example, if $X$ represents the number rolled on a six-sided die, then
$$
E X=\sum_{x=1}^{6} x P(X=x)=\sum_{x=1}^{6} x \frac{1}{6}=3.5
$$
is the expected value of $X$, which is the average of the integers $1-6$.