# The Expectation in terms of probability

let the independent random variables $X$ and $Y$ takes values in the range $\{1,...,n\}$

• Calculate the Expectation of $X$ in terms of $P(X \geqslant k)$.

Indeed,

note that $$\{X=k\}=\{X\geq k\}\cup \{X<k \}$$ then $$P\{X=k\}=P\{X\geq k\}+P\{X<k \}$$ $$P\{X=k\}=P\{X\geq k\}+1-P\{X \geq k+1\}$$ $$E[x]=\sum_{k=1}^{n}kP\{X\geq k\}+\sum_{k=1}^{n}k1-\sum_{k=1}^{n}kP\{X \geq k+1\}$$

• What you wrote doesn't make any sense at all. Suppose $n = 3$ and furthermore, suppose $\Pr[X = 1] = \Pr[X = 2] = \Pr[X = 3] = \frac{1}{3}$. Then $\Pr[X \ge 2] = \frac{2}{3}$ and $\Pr[X < 2] = \frac{1}{3}$, and $\Pr[X = 2] \ne \Pr[X \ge 2] + \Pr[X < 2]$. – heropup Jun 20 '14 at 6:15

\begin{align} p(1)+p(2)+p(3)+&... =P(X\ge 1)\\ p(2)+p(3)+&...=P(X\ge 2) \\ p(3)+&...=P(X\ge 3) \\ \end{align}\\

etc.