Distribution Functions

    

Random Variable: A random variable on a given probability space $(\Omega, \tilde{A}, P[.])$, denoted by $X$, is a function $X(.):\Omega \rightarrow R $   such that the set $\{w \in \Omega : X(w) \leq x \}$ as $\{X \leq x \}\in \tilde{A}$ for each $x\in R$.

It is customary to write the set $\{w \in \Omega: X(w)\leq x\} as \{X \leq x\}$.

Remarks 

1.  A random variable $X$ is a function such that $X(w)$ is a real number for each outcome $w$ of $\Omega$.
2. For each real number $x$, the set $\{w\in \Omega : X(w) \leq x\}$ is an event, since it belongs to $\tilde{A}$.
3. Every real-valued function defined on $\Omega$ may not be a random variable.

Example 1. The sample space of the experiment of tossing a coin is $\Omega = {H, T}$.

Define $X(.) : \Omega \rightarrow R$ as   $X(H)=1$ and $X(T)=0$.    .......(1)

Thus $X(.)$ associates a real number with each outcome of the experiment. 

Now we show that $A_x=\{w\in \Omega : X(w)\leq x\}$

We know $\tilde{A}=\{\phi, \{H\}, \{T\}, \Omega\}$

Using (1), we see that for $x<0, A_x = \phi \in \Omega$,

                                    for $0\leq x <1, A_x={T}\in \Omega$

                                    for $x\geq 1, A_x = {H, T}=\Omega \in \Omega$

Hence $A_x \in \Omega$ for each $x\in R$ and so $X(.)$ is a random variable.

Example 2. Consider a sample space  $(\Omega, \tilde{A}, P[.])$, where $\Omega = \{a, b, c, d\}$ and $\tilde{A}=\{\phi, \{a\}, \{b, c, d\}, \Omega\}$

Define $X(.) : \Omega \rightarrow R$ as $X(a)=0, X(b)=0$ and $X(c)=X(d)=1$ ,,... (1)

Thus $X(.)$ associates a real number with each outcome of the experiment. 

Now we show that $A_x=\{w\in \Omega : X(w)\leq x\}$

We know $\tilde{A}=\{\phi, \{a\}, \{b, c, d\}, \Omega\}$

Using (1), We see that for $x<0, A_x = \phi \in \Omega$

                                      for $0 \leq x < 1, A_x = \{a, b\} \notin \Omega$    

Hence $A_x \notin \Omega$ for each $x\in R $ and sso $X(.)$ is not a random variable.

Exercises:

1. The sample space of the experiment of the tossing a die in $\Omega=\{1, 2, 3, 4, 5, 6\}$. Then the variable $X(.) : \Omega \rightarrow R$ defined as $X(w)=w; w=1, 2, 3, 4, 5, 6$ is a random variable.

2. The constat function $X(.):\Omega \rightarrow R$ defined as $X(w)= k~ \forall~ w~ \in \Omega $, is a random variable.

3. Give two examples of random variables in which, one is random variable and other is not a random variable.

Indicator Function: If A is any subset of $\Omega$, then the indicator function of $A$, denoted by $I_A(.)$, is a function $I_A(.): \Omega \rightarrow \{0, 1\}$ such that 

$$I_A(w) = \left\{ \begin{array}{ll} 1, & if~ w\in A\\0, & if~ w\notin A \end{array}\right.$$

Examples:  

1. $I_{0,1}(x)=\left \{ \begin{array}{ll} 0& if~0<x<1 \\ 0, & otherwise \end{array} \right.$
2.  The function $f:R\rightarrow R$ defined by $f(x) = \left\{\begin{array}{ll}0, & for~ x\leq 0\\1, & for~ 0<x<1\\2, & for x\geq 1\end{array}\right.$

Can be written as $f(x) = I_{(0,1)}(x)+2I_{(1, \infty)}(x).$

Remark:  $I_A$ is a random variable.

Properties of Indicator Function:

1. $I_{\Omega}(w) = 1, I_{\phi}(w)=0$
2. $I_{\bar{A}}(w) = 1-I_A(w)$ for each $A\in \tilde{A}$.
3. $I_{A\cup B}(w) = max{I_A(w), I_B(w)}$
4. $I_A^2(w) = I_A (w)$ for $A \in \tilde{A}$

Cumulative Distribution Function: The cumulative distribution function or simple distribution function of a random variable X, denoted by $F_X(.)$, is defined as a function $ F_X(.):R\rightarrow [0,1]$ such that

$F_X(x)=P[X\leq x], \forall x\in R$. This implies that $0\leq F_X(x) \leq 1$.

Recall that $\{X\leq x\}=\{w:X(w) \leq x\}$

This implies that $0\leq F_x(x)\leq 1$. Sometimes we write $F(x)$ instead of $F_X(x)$.

Properties of Distribution Function:

    If $F_X(.)$ is the distribution function of the random variable $X$ and $a < b,$ then

1. $F_X(a) \leq F_X(b)~for~ a<b \Rightarrow$ Distribution function is monotonically increasing function.
2. $F_X(.)$ is continuous from the right i.e., $lim_{n\rightarrow} F_X(x+h)=F_X(x)$.
3. $F_X(-\infty) = lim_{x\rightarrow -\infty} F_X(x)=0$ and $F_X(\infty) = lim_{x\rightarrow\infty} F_X(x)=1.$

Discrete random variable: A random variable $X$ is said to be discrete if the range of $X$ is countable i.e. $X$ takes the values  $x_1, x_2, x_3,...x_n,...$. The countable values of  $X$ are called the mass point of $X$. The distribution function of a discrete random variable $X$ is called a discrete distribution function.

Discrete Density Function: If $X$ is a discrete random variable with distinct values $x_1, x_2,..., x_n,....$ then the function $f_X(.):R\rightarrow [0, 1],$ satisfying

$$f_X(x) = \left\{ \begin{array}{ll}P[X=x_i], & if~ x=x_i, i=1, 2,..., n\\ 0, & if~x\neq x_i, \forall i \in N\end{array}\right.$$

is called the discrete density function of $X$ or prabability mass function or probability function of $X$. The value $f_X(x_i)$ or $f(x_i)$ or is called the mass associated with the mass point $x_i$ such that

1. $f(x_i)>0~ for~ i=1, 2,....., $
2. $f(x)=0~for~x\neq x_i, i=1, 2,....$
3. $\sum_{n=1}^{\infty} x_n = 1$ 

Example 1. A random variable has the following probability distribution:

$x$	0	1	2	3	4	5	6	7	8
$p(x)$	$k$	$3k$	$5k$	$7k$	$9k$	$11k$	$13k$	$15k$	$17k$

1. Determine the value of $k$.
2. Find $P[X < 4], P[X\geq 5], P[0<X<4]$
3. Find the distribution function.
4. Find the smallest value of $x$ for which $P[X \leq x]>\frac{1}{2}$.

Solution: 

1. Here $f(x)=p(x) \Rightarrow \sum p(x) = 1 \Rightarrow [k+3k+5k+7k+9k+11k+13k+15k+17k]=1$  $\Rightarrow k=\frac{1}{81}$.
2. $P[X<4]=P[X=0]+P[X=1]+P[X=2]+P[X=3]=k+3k+5k+7k=\frac{16}{81}$                                     $P[X\geq 5] = P[X=5] + P[X= 6]+ P[X=7] +P[X=8]=11k+13k+15k+17k = \frac{56}{81}$             $P[0<X<4]=P[X=1] + P[X=2]+P[X=3] = 3k +5k+7k = \frac{15}{81}$
3. If $F(x)$ is the distribution function then

$x$	0	1	2	3	4	5	6	7	8
$f(x)=p(x)$	1/81	3/81	5/81	7/81	9/81	11/81	13/81	15/81	17/81
$F(x)=P[X\leq x]$	1/81	4/81	1/9	16/81	25/81	4/9	49/81	64/81	1

        4. As we know that $\frac{4}{9} < \frac{1}{2}$ and $\frac{49}{81}>\frac{1}{2}$, therefore from the distribution table, we find that $F(x) = P[X\leq x] >\frac{1}{2}$ for $x=6$.

Exercises:

1.  Let $p(x)$ be the probability function of a discrete random variable $X$ which assumes the values $x_1, x_2, x_3, x_4$   such that $2p(x_1)=3p(x_2)=p(x_3)=5p(x_4)$. Find probability distribution and cumulative distribution of $X$.

2. Let $X$ be a random variable such that $P[X=-2]=P[X=-1], P[X=2]=P[X=1]$ and $P[X>0]=P[X<0]=P[X=0]$. Obtain the probability mass function and its distribution function.

3. A random variable $X$ can take all non-negative integral values, and the probability that $X$ takes the value $r$ is proportional to $\alpha^r$ where $(0<\alpha <1)$. Find $P[X=0]$.

4. A random variable $X$ takes values 0, 1, 2, ....., with probability proportional to $(x+1)(\frac{1}{5})^x$. Find the probability that $X\leq 5.$

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