5.3: Expected value
- Page ID
- 24550
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
Definition:
For a probability distribution defined by \(P(X =x) \), we define the expectation of the random variable \(X \) as
\[ \begin{align*} E(X) &= \sum_{i=1}^{i=n} x_i P(X=x_i) \\[4pt] &= x_1 P(X=x_2) + x_2 P(X=x_2)+ \cdots+ x_n P(X=x_n) \end{align*}\]
where \(x_i\) represents the observed outcome and \(P(X=x_i)\) is the probability of the outcome occurring.
Note: Every discrete distribution must satisfy
\(P(X=x_i)\ge 0\) and \(\sum_iP(X=x_i)=1.\)
The “expected value of \(X\)” can be interpreted as the mean value of \(X.\)
The expectation values can be considered in two ways.
1. Long-run average
This is the measure one would see if the experiment was repeated a large number of times, namely \(E(X)= np\), where \(n\) is the number of times the experiment occurred and \(p\) is the probability for the event to occur.
Example \(\PageIndex{1}\)
If we tossed a coin \(1500 \) times, and the random variable \(X,\) represents the number of heads observed, we would expect \(750 \) heads, that is \(E(X) = 750\).
2. Probability-weighted average
This measure accounts for the relative probabilities of each observed outcome.
Example \(\PageIndex{2}\)
For the probability distribution with random variable X defined by
| \(x\) | \(2\) | \(3\) | \(4\) | \(5\) |
| \(P(X=x)\) | \(\dfrac{1}{6}\) | \(\dfrac{1}{6}\) | \(\dfrac{1}{6}\) | \(\dfrac{1}{2}\) |
\(E(X)= \sum_{i=1}^{i=4} x_i P(X=x_i) = 2 \dfrac{1}{6} +3 \dfrac{1}{6}+4 \dfrac{1}{6} +5 \dfrac{1}{2}= \dfrac{24}{6}=4\).
Thus \(X\) has a mean value of \( 4\).
Example \(\PageIndex{3}\):
We toss four coins at the same time, then the probability of getting \(X\) number of tails:
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | Total |
| \(P(X=x)\) | \(\dfrac{1}{16}\) | \(\dfrac{4}{16}\) | \(\dfrac{6}{16}\) | \(\dfrac{4}{16}\) | \(\dfrac{1}{16}\) |
Then, the expected value is
\(E(X)= \sum_{i=0}^{i=4} x_i P(X=x_i) = 0 \dfrac{1}{16} +1 \dfrac{4}{16}+2 \dfrac{6}{16}+3 \dfrac{4}{16}+ 4 \dfrac{1}{16}= \dfrac{32}{16}=2\).
Therefore, in the long run, we expect to get \(2\) tails when tossing \(94\) coins simultaneously.