Example 23: A discrete random variable X has the following distribution function : 0, for x < 1 1/3, for 1 < x < 4 F(x)=1/2, for 4 ≤ x < 6 5/6, for 6
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Example 23: A discrete random variable X has the following distribution function : 0, for x < 1 1/3, for 1 < x < 4 F(x)=1/2, for 4 ≤ x < 6 5/6, for 6
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Answer:
The distribution function you provided defines the cumulative distribution function (CDF) of the random variable X. To find the probability mass function (PMF) for this discrete random variable, we can calculate the probability of X taking specific values.
First, let's find the probability mass function (PMF) for X:
P(X = 1) = F(1) - F(1-) = 1/3 - 0 = 1/3
P(X = 2) = F(2) - F(2-) = F(2) - F(1-) = 1/3 - 0 = 1/3
P(X = 3) = F(3) - F(3-) = F(3) - F(2-) = 1/3 - 1/3 = 0
P(X = 4) = F(4) - F(4-) = F(4) - F(3-) = 1/2 - 1/3 = 1/6
P(X = 5) = F(5) - F(5-) = F(5) - F(4-) = 1/2 - 1/2 = 0
P(X = 6) = F(6) - F(6-) = F(6) - F(5-) = 5/6 - 1/2 = 1/3
So, the probability mass function (PMF) for the random variable X is as follows:
P(X = 1) = 1/3
P(X = 2) = 1/3
P(X = 3) = 0
P(X = 4) = 1/6
P(X = 5) = 0
P(X = 6) = 1/3
These probabilities describe the likelihood of the random variable X taking on each specific value.