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Let $F: \mathbb{R}\to \mathbb{R}$. Then $F$ is a \emph{distribution function} if
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Let $F: \mathbf{R}\to \mathbf{R}$. Then $F$ is a \emph{distribution function} if
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| \begin{enumerate} |
\begin{enumerate} |
| \item |
\item |
| $F$ is nondecreasing, |
$F$ is nondecreasing, |
| \item |
\item |
| $F$ is continuous from the right, |
$F$ is continuous from the right, |
| \item |
\item |
| $\lim_{x \rightarrow -\infty} F(x) = 0$, and $\lim_{x \rightarrow \infty} F(x) = 1$. |
$\lim_{x \rightarrow -\infty} F(x) = 0$, and $\lim_{x \rightarrow \infty} F(x) = 1$. |
| \end{enumerate} |
\end{enumerate} |
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| As an example, suppose that $\Omega = \mathbb{R}$ and that $\mathcal{B}$ |
As an example, suppose that $\Omega = \mathbb{R}$ and that $\mathcal{B}$ |
| is the $\sigma$-algebra of Borel subsets of $\mathbb{R}$. |
is the $\sigma$-algebra of Borel subsets of $\mathbb{R}$. |
| Let $P$ be a probability measure on $(\Omega, \mathcal{B})$. |
Let $P$ be a probability measure on $(\Omega, \mathcal{B})$. |
| Define $F$ by |
Define $F$ by |
| $$ |
$$ |
| F(x) = P((-\infty, x]). |
F(x) = P((-\infty, x]). |
| $$ |
$$ |
| This particular $F$ is called the \emph{distribution function} of $P$. It is |
This particular $F$ is called the \emph{distribution function} of $P$. It is |
| easy to verify that 1,2, and 3 hold for this $F$. |
easy to verify that 1,2, and 3 hold for this $F$. |
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| In fact, every distribution function is the distribution function of some |
In fact, every distribution function is the distribution function of some |
| probability measure on the Borel subsets of $\mathbb{R}$. To see this, |
probability measure on the Borel subsets of $\mathbb{R}$. To see this, |
| suppose that $F$ is a distribution function. We can define $P$ on a single half-open |
suppose that $F$ is a distribution function. We can define $P$ on a single half-open |
| interval by |
interval by |
| $$ |
$$ |
| P((a,b]) = F(b) - F(a) |
P((a,b]) = F(b) - F(a) |
| $$ |
$$ |
| and extend $P$ to unions of disjoint intervals by |
and extend $P$ to unions of disjoint intervals by |
| $$ |
$$ |
| P( \cup_{i=1}^\infty (a_i, b_i])= \sum_{i=1}^\infty P((a_i, b_i]). |
P( \cup_{i=1}^\infty (a_i, b_i])= \sum_{i=1}^\infty P((a_i, b_i]). |
| $$ |
$$ |
| and then further extend $P$ to all the Borel subsets of $\mathbb{R}$. |
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| It is clear that the distribution function of $P$ is $F$. |
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| \subsection{Random Variables} |
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| Suppose that $(\Omega, \mathcal{B}, P)$ is a probability space and |
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| $X: \Omega \to \mathbb{R}$ is a random variable. Then there is an |
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| \emph{induced} probability measure $P_X$ on $\mathbb{R}$ defined as |
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| follows: \\ |
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| $$ |
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| P_X(E) = P(X^{-1}(E)) |
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| $$ |
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| for every Borel subset $E$ of $\mathbb{R}$. $P_X$ is called the |
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| \emph{distribution} of $X$. The \emph{distribution function} |
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| of $X$ is |
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| $$ |
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| F_X(x) = P(\omega | X(\omega) \leq x). |
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| $$ |
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| Claim: $F_X$ = the distribution function of $P_X$. |
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| \begin{eqnarray*} |
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| F_X(x) &=& P(\omega | X(\omega) \leq x) \\ |
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| &=& P(X^{-1}((-\infty, x]) \\ |
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| &=& P_X((-\infty, x]) \\ |
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| &=& F(x). |
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| \end{eqnarray*} |
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and then further extend $P$ to all the Borel subsets of $\mathbb{R}$. |
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It is clear that the distribution function of $P$ is $F$. |
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\subsection{Random Variables} |
| \subsection{Density Functions} |
\subsection{Density Functions} |
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| If $X$ is a discrete random variable, we have then |
If $X$ is a discrete random variable, we have then |
| $$F(x)=\sum_{x_j\leq x} f(x_j)$$ |
$$F(x)=\sum_{x_j\leq x} f(x_j)$$ |
| and if $X$ is a continuous random variable then |
and if $X$ is a continuous random variable then |
| $$F(x)=\int_{-\infty}^x f(t) dt$$ |
$$F(x)=\int_{-\infty}^x f(t) dt$$ |
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| Due to the properties of integrals and summations, we can use the distribution function to calculate the probability of $X$ being on a given interval: |
Due to the properties of integrals and summations, we can use the distribution function to calculate the probability of $X$ being on a given interval: |
| $$P[a<X\leq b]=F(b)-F(a).$$ |
$$P[a<X\leq b]=F(b)-F(a).$$ |
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| Two special cases arise:\\ |
Two special cases arise:\\ |
| From the definition: $P[X\leq b]=F(b)$, and $P[a<X]=1-F(a)$ since the complement of an interval $(a,\infty)$ is an interval $(-\infty,a]$ and together they cover the whole sample space. |
From the definition: $P[X\leq b]=F(b)$, and $P[a<X]=1-F(a)$ since the complement of an interval $(a,\infty)$ is an interval $(-\infty,a]$ and together they cover the whole sample space. |
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| For the continuous case, we have a relationship linking the density function and the distribution function: |
For the continuous case, we have a relationship linking the density function and the distribution function: |
| $$F'(x)=f(x).$$ |
$$F'(x)=f(x).$$ |
