Discrete random variables- Cumulative Distribution Function
QUESTION 1:
The discrete random variable X has the following probability distribution function.
f(x)={■(1/6&x=0,2@1/3&x=1,3@0&otherwise)}
(a) Find the cumulative distribution function F(x)
(b) P(X≤2)
(c) P(X<1)
(d) P(0<X≤3)
(e) P(X>2)
(f) P(X<-1)
QUESTION 2:
The discrete random variable X has the following probability distribution function.
f(x)={■(1/8&x=0,3@3/8&x=1,2@0&otherwise)}
(a) Find the cumulative distribution function F(x)
(b) P(X≤-0.5)
(c) P(X≤2.5)
(d) P(X≤4)
QUESTION 3
The cumulative distribution function F(x) of a discrete random variable X is given by:
F(x)=kx,x=1.2,3,4
Find
(a) the value of the constant k
(b) the probability distribution of X
ANSWER 1:
(a) F(x) = {■(0&x<0@1/6&x=0@2/3&x=1@1&x=2@1&x>2)}
(b) P(X≤2) = 1
(c) P(X<1) = 2/3
(d) P(0<X≤3) = 1
(e) P(X>2) = 0
(f) P(X<-1) = 0
ANSWER 2:
(a) F(x) = {■(0&x<0@1/8&x=0@4/8&x=1@7/8&x=2@1&x>2)}
(b) P(X≤-0.5) = 0
(c) P(X≤2.5) = 1
(d) P(X≤4) = 1
ANSWER 3:
(a) The value of the constant k = 1/4
(b) The probability distribution of X = {■(1/4&x=1@1/2&x=2@3/4&x=3@1&x=4)}
QUESTION 1:
(a) To find the cumulative distribution function (CDF), we need to sum up the probabilities of all the outcomes up to a certain value of x.
For x = 0, F(0) = P(X ≤ 0) = f(0) = 1/6.
For x = 1, F(1) = P(X ≤ 1) = f(0) + f(1) = 1/6 + 1/3 = 1/2.
For x = 2, F(2) = P(X ≤ 2) = f(0) + f(1) + f(2) = 1/6 + 1/3 + 1/6 = 2/3.
For x = 3, F(3) = P(X ≤ 3) = f(0) + f(1) + f(2) + f(3) = 1/6 + 1/3 + 1/6 + 0 = 5/6.
For all other values of x, F(x) = 1 since the probability is 0.
Therefore, the CDF is:
F(x) = {■(1/6& x=0@ 1/2& x=1@ 2/3& x=2@ 5/6& x=3@ 1& otherwise)}
(b) To find P(X ≤ 2), we can use the CDF:
P(X ≤ 2) = F(2) = 2/3.
(c) To find P(X < 1), we can subtract the probability of X = 1 from P(X ≤ 1):
P(X < 1) = P(X ≤ 1) - P(X = 1) = F(1) - f(1) = 1/2 - 1/3 = 1/6.
(d) To find P(0 < X ≤ 3), we can subtract the probability of X = 0 from P(X ≤ 3):
P(0 < X ≤ 3) = P(X ≤ 3) - P(X = 0) = F(3) - f(0) = 5/6 - 1/6 = 2/3.
(e) To find P(X > 2), we can subtract P(X ≤ 2) from 1:
P(X > 2) = 1 - P(X ≤ 2) = 1 - F(2) = 1 - 2/3 = 1/3.
(f) Since there is no value less than 0, P(X < -1) = 0.
QUESTION 2:
(a) Similarly, let's find the cumulative distribution function (CDF) for this probability distribution:
For x = 0, F(0) = P(X ≤ 0) = f(0) = 1/8.
For x = 1, F(1) = P(X ≤ 1) = f(0) + f(1) = 1/8 + 3/8 = 1/2.
For x = 2, F(2) = P(X ≤ 2) = f(0) + f(1) + f(2) = 1/8 + 3/8 + 0 = 1/2.
For x = 3, F(3) = P(X ≤ 3) = f(0) + f(1) + f(2) + f(3) = 1/8 + 3/8 + 0 + 0 = 1/2.
For all other values of x, F(x) = 1 since the probability is 0.
Therefore, the CDF is:
F(x) = {■(1/8& x=0@ 1/2& x=1,2,3@ 1& otherwise)}
(b) Since there is no x less than 0, P(X ≤ -0.5) = 0.
(c) To find P(X ≤ 2.5), we can use the CDF:
P(X ≤ 2.5) = F(2) = 1/2.
(d) Since there is no x greater than 3, P(X ≤ 4) = 1/2.
QUESTION 3:
(a) To find the value of the constant k, we need to use the fact that the CDF is a non-decreasing function. Therefore, for each value of x, F(x) should be greater than or equal to F(x-1).
From the given CDF, we have F(1) = k*1 = k,
F(2) = k*2 = 2k,
F(3) = k*3 = 3k,
F(4) = k*4 = 4k.
Since F(2) ≤ F(3) and F(3) ≤ F(4), it implies 2k ≤ 3k and 3k ≤ 4k.
By canceling k from both sides, we get 2 ≤ 3 and 3 ≤ 4.
Since both inequalities hold true, we can conclude that k can be any positive number.
(b) Since the CDF only provides information about probabilities up to integer values of x, we cannot determine the probability distribution of X without further information.
To answer these questions, we need to understand the concepts of cumulative distribution function (CDF) and probability distribution function (PDF) for discrete random variables.
1. CDF:
The cumulative distribution function F(x) gives the probability that a random variable X is less than or equal to a particular value x. It is obtained by summing up the probabilities of all values less than or equal to x.
2. PDF:
The probability distribution function f(x) gives the probability of each possible value of a discrete random variable X. It is often provided as a table or a function.
Now, let's solve the given questions step by step.
QUESTION 1:
(a) To find the cumulative distribution function F(x), we will sum up the probabilities of all values less than or equal to x.
F(x) = P(X ≤ x)
For x = 0: F(0) = P(X ≤ 0) = f(0) = 1/6
For x = 1: F(1) = P(X ≤ 1) = f(0) + f(1) = 1/6 + 1/3 = 1/2
For x = 2: F(2) = P(X ≤ 2) = f(0) + f(1) + f(2) = 1/6 + 1/3 = 1/2
For x = 3: F(3) = P(X ≤ 3) = f(0) + f(1) + f(2) = 1/6 + 1/3 = 1/2
For x > 3: F(x) = P(X ≤ 3) = f(0) + f(1) + f(2) = 1/6 + 1/3 + 0 = 1/2
Thus, the cumulative distribution function F(x) is:
F(x) = { 1/6 for x = 0
1/2 for x = 1, 2, 3
0 for x > 3 }
(b) P(X ≤ 2) = F(2) = 1/2
(c) P(X < 1) = F(0) = 1/6
(d) P(0 < X ≤ 3) = F(3) - F(0) = 1/2 - 1/6 = 1/3
(e) P(X > 2) = 1 - P(X ≤ 2) = 1 - F(2) = 1 - 1/2 = 1/2
(f) P(X < -1) = 0 (since the random variable X cannot take negative values)
QUESTION 2:
(a) Following a similar approach as Question 1, we calculate the cumulative distribution function F(x).
F(x) = { 1/8 for x = 0
5/8 for x = 1
1 for x = 2
0 for x > 2 }
(b) P(X ≤ -0.5) = 0 (since the random variable X cannot take negative values)
(c) P(X ≤ 2.5) = F(2) = 1
(d) P(X ≤ 4) = 1 (since the random variable X's maximum value is 2)
QUESTION 3:
(a) The given cumulative distribution function F(x) can be represented as a piecewise function for the corresponding probabilities.
F(1) = k
F(2) = 2k
F(3) = 3k
F(4) = 4k
Since F(x) is the cumulative distribution function, it should satisfy the properties:
F(1) = f(1)
F(2) = f(1) + f(2)
F(3) = f(1) + f(2) + f(3)
F(4) = f(1) + f(2) + f(3) + f(4)
From the above equations, we can see that f(x) = F(x) - F(x-1)
f(1) = F(1) - F(0) = k
f(2) = F(2) - F(1) = 2k - k = k
f(3) = F(3) - F(2) = 3k - 2k = k
f(4) = F(4) - F(3) = 4k - 3k = k
Thus, the probability distribution of X is:
f(x) = { k for x = 1, 2, 3, 4
0 for all other values of x }
(b) To find the value of the constant k, we can use the fact that the sum of all probabilities should be equal to 1.
1 = f(1) + f(2) + f(3) + f(4) = 4k
Hence, k = 1/4.
Therefore, the value of the constant k is 1/4, and the probability distribution of X is:
f(x) = { 1/4 for x = 1, 2, 3, 4
0 for all other values of x }
I hope this helps! Let me know if you have any further questions.