Question

andom variable X has CDF given as follows. Find the PMF of X.Find the PMF of X, and confirm that it is a valid PMF.

Answer 1

Complete Question

Random variable X has CDF given as follows

        $F(x)_x = \left \{ {{{{{{{ 0 \ \ \ \ \ \ \ \ \ \ x < 0} \atop {\frac{8}{33}} \ \ \ \ \ \ 0 \le x \le 1} \atop{ \frac{5}{11} }\ \ \ \ \ 1 \le x \le 2 } \atop {\frac{7}{11} } \ \ \ \ \ \ 2 \le x \le 3} \atop{ \frac{26}{33} }\ \ \ \ \ 3 \le x \le 4 }\atop {\frac{10}{11} } \ \ \ \ \ \ 4 \le x \le 5} \atop{ 1}\ \ \ \ \ \ \ \ 5 \le x } \right.$

Find the PMF of X.Find the PMF of X, and confirm that it is a valid PMF.  

Answer:

Generally the PMF is

      X               0        1       2      3         4       5  

    P(X = x)      $\frac{8}{33}$        $\frac{7}{33}$      $\frac{2}{11}$     $\frac{5}{33}$        $\frac{4}{33}$     $\frac{1}{11}$

Generally to prove the validity of the PMF we add the probabilities , if it is equal to  1 then the PMF is valid otherwise , the PMF is not valid

So

      $\sum P(X = x) = \frac{8}{33} + \frac{7}{33} + \frac{2}{11} + \frac{5}{33} + \frac{4}{33}+ \frac{1}{11}$

      $\sum P(X = x) =1$

Hence the PMF is valid

Step-by-step explanation:

Generally the probability of  X = 0  is evaluated as

       $P(0) = F(0) \ \ (for \ x \le 0) - F(0) \ ( for \ x < 0)$

=>    $P(0) = \frac{8}{33} - 0$

=>    $P(0) = \frac{8}{33}$

Generally the probability of  X = 1  is evaluated as

       $P(1) = F(1) \ \ (for \ 1 \le 0) - F(1) \ ( for \ x < 1)$

=>    $P(1) = \frac{5}{11} - \frac{8}{33}$

=>    $P(1) = \frac{7}{33}$

Generally the probability of  X = 2  is evaluated as

       $P(2) = F(2) \ \ (for \ 2 \le x ) - F(2) \ ( for \ x < 2)$

=>    $P(2) = \frac{7}{11} - \frac{5}{11}$

=>    $P(2) = \frac{2}{11}$

Generally the probability of  X = 3  is evaluated as

       $P(3) = F(3) \ \ (for \ 3 \le x ) - F(3) \ ( for \ x < 3)$

=>    $P(3) = \frac{26}{33} - \frac{7}{11}$

=>    $P(3) = \frac{5}{33}$

Generally the probability of  X = 4  is evaluated as

       $P(4) = F(4) \ \ (for \ 4 \le x ) - F(4) \ ( for \ x < 4)$

=>    $P(4) = \frac{10}{11} - \frac{26}{33}$

=>    $P(4) = \frac{4}{33}$

Generally the probability of  X = 5  is evaluated as

       $P(5) = F(5) \ \ (for \ 5 \le x ) - F(5) \ ( for \ x < 5)$

=>    $P(5) = 1- \frac{10}{11}$

=>    $P(5) = \frac{1}{11}$

Generally the PMF is

      X               0        1       2      3         4       5  

    P(X = x)      $\frac{8}{33}$        $\frac{7}{33}$      $\frac{2}{11}$     $\frac{5}{33}$        $\frac{4}{33}$     $\frac{1}{11}$

Generally to prove the validity of the PMF we add the probabilities , if it is equal to  1 then the PMF is valid otherwise , the PMF is not valid

So

      $\sum P(X = x) = \frac{8}{33} + \frac{7}{33} + \frac{2}{11} + \frac{5}{33} + \frac{4}{33}+ \frac{1}{11}$

      $\sum P(X = x) =1$

Hence the PMF is valid