GO Classes DPP | GATE CS, DA | Probability & Statistics | Probability Mass Function

Answer 1

Finding the value of $c$ :

For any valid probability mass function (PMF), the sum of all probabilities must equal $1$.

$$ \sum p_X(x) = 1 $$

$$ \Rightarrow c + 2c + \frac{1}{6} + 3c + \frac{1}{3} = 1 $$

$$ \Rightarrow c = \frac{1}{12} $$

Now we can write the complete PMF:

• $P(X = -1) = \frac{1}{12}$

• $P(X = 1) = 2(\frac{1}{12}) = \frac{2}{12}$

• $P(X = 4) = \frac{1}{6} = \frac{2}{12}$

• $P(X = 6) = 3(\frac{1}{12}) = \frac{3}{12}$

• $P(X = 8) = \frac{1}{3} = \frac{4}{12}$

 

Determining the Cumulative Distribution Function $F_X(x)$

The CDF is defined as $F_X(x) = P(X \le x)$. We calculate it by maintaining a running total of the probabilities:

• For $x < -1$, $F_X(x) = 0$

• For $-1 \le x < 1$, $F_X(x) = \frac{1}{12}$

• For $1 \le x < 4$, $F_X(x) = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4}$

• For $4 \le x < 6$, $F_X(x) = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$

• For $6 \le x < 8$, $F_X(x) = \frac{5}{12} + \frac{3}{12} = \frac{8}{12} = \frac{2}{3}$

• For $x \ge 8$, $F_X(x) = \frac{8}{12} + \frac{4}{12} = \frac{12}{12} = 1$

 

Calculating the Expected Value $E[X]$

The expected value is the sum of each value multiplied by its probability:

$$ E[X] = \sum [x \cdot P(X = x)] $$

$$ E[X] = (-1)\left(\frac{1}{12}\right) + (1)\left(\frac{2}{12}\right) + (4)\left(\frac{2}{12}\right) + (6)\left(\frac{3}{12}\right) + (8)\left(\frac{4}{12}\right) $$

$$ E[X] = \frac{-1 + 2 + 8 + 18 + 32}{12} = \frac{59}{12} $$

 

Now, We need to find $E[X] + F_X(5) - 4F_X(0)$ : 

First, find the required CDF values based on the intervals we defined earlier while finding the CDF:

• $F_X(5)$: Since $5$ falls in the interval $4 \le x < 6$, $F_X(5) = \frac{5}{12}$.

• $F_X(0)$: Since $0$ falls in the interval $-1 \le x < 1$, $F_X(0) = \frac{1}{12}$.

 

$$ \Rightarrow E[X] + F_X(5) - 4F_X(0) = \frac{59}{12} + \frac{5}{12} - 4\left(\frac{1}{12}\right) = \boxed{5} $$