题目

Problem

The discrete random variable $X$ has the following probability distribution

$x$	$-2$	$-1$	$1$	$3$	$4$
$P(X=x)$	0.15	$a$	$b$	$c$	0.15

where $a$ , $b$ and $c$ are probabilities.

The mean value of $X$ is 1 and $F(1)=0.63$

Find the value of $a$ , the value of $b$ and the value of $c$ .

(5)

解答

展开

这里有三个未知数，所以需要三个条件：总概率为 1， $F(1)=P(X\leq1)$ ，以及 $E(X)=1$ 。

展开

Since the probabilities sum to 1,

\begin{align*} a+b+c+0.15+0.15=1. \end{align*}

\begin{align*} a+b+c=0.70. \end{align*}

Also,

\begin{align*} F(1)=P(X\leq1)=0.15+a+b. \end{align*}

Given $F(1)=0.63$ ,

\begin{align*} a+b=0.48. \end{align*}

Therefore

\begin{align*} c=0.70-0.48=0.22. \end{align*}

Using $E(X)=1$ ,

\begin{align*} -2(0.15)-a+b+3c+4(0.15)=1. \end{align*}

Substitute $c=0.22$ :

\begin{align*} -0.30-a+b+0.66+0.60=1. \end{align*}

Thus

\begin{align*} -a+b=0.04. \end{align*}

Together with $a+b=0.48$ ,

\begin{align*} 2b=0.52, \end{align*}

\begin{align*} b=0.26. \end{align*}

Hence

\begin{align*} a=0.48-0.26=0.22. \end{align*}

Therefore

\begin{align*} a=0.22,\qquad b=0.26,\qquad c=0.22. \end{align*}