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Cumulative Distribution Function (CDF) of any random variable, say ‘X’, that is evaluated at x (any point), is the probability function that ‘X’ will take a value equal to or less than x. A variable that defines the possible outcome values of any phenomenon is called a random variable. Cumulative Distribution Function is defined for both random and discrete variables. In case, a random variable is above a particular level then, CDF is known as the Complementary Cumulative Distribution Function (CCDF) or tail distribution.
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Key Takeaways: Cumulative Distribution Function, random variable, probability, probability distribution, cumulative frequency, frequency distribution
Cumulative Distribution Function
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Cumulative Distribution Function (CDF) of any random variable, say ‘X’, that is evaluated at x, is the probability function that X will take a value equal to or less than x.
Graph of Cumulative Distributive Function
CDF is used to find the cumulative probability for a given value. It is used to calculate the probability of a random variable and also to compare the probability of variables under a given condition. For continuous distribution functions, CDF gives the area under the probability density function.
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Formula of Cumulative Distribution Function
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The Cumulative Distribution Function for a discrete random variable is defined as
FX(x) = P(X ≤ x)
Where X is the probability that takes a value equal to or less than x and it lies between the interval (a,b], a<b.
Thus, the probability with the interval is given by,
P(a < X ≤ b) = FX(b) - FX(a)
Note here that the ‘≤’ sign used here is not necessarily used all the time, but it can be used for discrete distributions.
For a continuous random variable, the CDF is defined as:
FX(x) = -∞∫x fX(t)dt
Where X is expressed in terms of integration of its probability density function fx.
If the distribution of the random variable X has the discrete component at value b, then it is given by:
P(X = b) = FX(b) - limx→b FX(x)
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Properties of Cumulative Distribution Function
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The Cumulative Distribution Function FX(x) of the variable has some important properties. If any of the given properties are fulfilled, the function can be said as the cumulative distribution function of that variable.
- Every CDF function is a non-increasing and right continuous function. In other words, Limx\(\rightarrow\)-∞ FX(x) = 0 and limx\(\rightarrow\)+∞ FX(x) = 1.
- If ‘X’ is defined as a discrete random variable then its value is x1, x2, x3, …… etc, and the probability Pi = p(xi). Thus, the CDF of the variable is discontinuous at x1, x2, x3,... FX(x) = P(X ≤ x) = ∑xi ≤ x. P(X = xi) = ∑i ≤ x p(xi).
- If the CDF of a real-valued function is said to be continuous, then ‘X’ is called a continuous random variable Fx(b) - Fx(a) = P(a < X ≤ b) = ∫ab fX(x) dx.
The function fX is a derivative of FX that is defined as the probability density function of X.
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| Independent Events in Probability | Multiplication Theory on Probability | Conditional Probability Formula |
Derived Functions
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- Complementary Cumulative Distribution Function (CCDF): Also known as tail distribution or exceedance. It is defined as, FX(x) = P(X>x) = 1 - FX(x).
- Folded Cumulative Distribution: After plotting the cumulative distributive function, it resembles the ‘S’ shape, this shape is known as Folded Cumulative Distribution (FCD) or mountain plot.
- Inverse Distribution Function: The Inverse Distribution Function or the quantile function is defined when the Cumulative Distribution Function (CDF) is continuous and increasing.
- Empirical Distribution Function: The estimation of cumulative distribution function for the points generated on the sample is called empirical distribution function.
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Cumulative Frequency Distribution
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Cumulative Frequency Distribution is defined as a set of data written in tabular or graphical form, representing the frequency of observations in a given interval. For cumulative frequency, the total number of observations that occurred beyond specified observation is calculated.
Applications of Cumulative Distribution Function
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In statistics and probability, the Cumulative Frequency Distribution is used in two ways:
- To find the frequency of occurrence of values using cumulative frequency analysis.
- Derive statistical properties with the help of empirical distribution functions.
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Things to Remember
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- Cumulative Distribution Function (CDF) of any random variable, say ‘X’, that is evaluated at x (any point), is the probability function that ‘X’ will take a value equal to or less than x.
- The estimation of cumulative distribution function for the points generated on the sample is called empirical distribution function.
- Complementary Cumulative Distribution Function (CCDF): Also known as tail distribution or exceedance. It is defined as, FX(x) = P(X>x) = 1 - FX(x).
- CDF is used to find the cumulative probability for a given value. It is used to calculate the probability of a random variable and also to compare the probability of variables under a given condition.
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Sample Questions
Ques: Find a probability distribution formed by rolling six-faced fair dice. Take X as a random variable. Also, find the probability range. (3 Marks)
Ans. The probability of getting at outcome by rolling a fair dice is given by:
Probability of getting 1 = P(X ≤ 1) = 1/6
Probability of getting 2 = P(X ≤ 2) = 2/6
Probability of getting 3 = P(X ≤ 3) = 3/6
Probability of getting 4 = P(X ≤ 4) = 4/6
Probability of getting 5 = P(X ≤ 5) = 5/6
Probability of getting 6 = P(X ≤ 6) = 6/6
From the above discussion, it is noted the value of probability always lies between 0 and 1. It can be represented as: 0 ≤ P(X ≤ x) ≤ 1.
Ques: The variable X is given by
f(x) = 2x3, 0 < x < 1
What is the Cumulative Distribution Function F(x)? (3 Marks)
Ans. The Cumulative Distribution Function of the variable is denoted by F(x) and can be written as
F(x) = 0∫1f(x)dx….(1)
Let consider f(t) = 2t3
We have,
F(x) = 0∫x 2t3dt = 2[t4/4]0x = x4/2 , 0<x<1
Therefore,
F(x) = 0, x ≤ 0
x4/2 , 0<x<1
Ques: Consider the random variable X that takes values -1 and +1 with probability 0.45 and 0.55 respectively. Find the values of the CDF F(x) at x = -1 and x = +1. (3 Marks)
Ans. Given that,
P(-1) = 0.45 and
P(1) = 0.55
So, Probability at all other points will be zero as the sum of probabilities must be 1.
Let P is the probability, F(x) is the Cumulative Distribution Function and X is a random variable.
Thus,
F(x) = P[X ≤ x]
F(-1) = P(X ≤ -1) = 0.45
and
F(1) = P[X ≤ 1]
= P(x = -1) + P(x = 1)
= 0.45 + 0.55
= 1
So, the CDF value at x = -1 is 0.45 and at x = 1 it is 1
Ques: Let X is a random variable giving the number of HEADS - number of TAILS in 2 tosses of a fair coin. Find the probability distribution of X. (5 Marks)
Ans. Sample Point (SS) = {HH, HT, TH, TT}
X = random variable
X = no. of Heads - no. of Tails
P(X = x) = Probability Distribution
P(X ≤ x) = Cumulative Distribution Function (CDF)
| Sample Point (SS) | X = no. Heads - no. of Tails |
|---|---|
| HH | 2 - 0 = 2 |
| HT | 1 - 1 = 0 |
| TH | 1 -1 = 0 |
| TT | 0 - 2 = -2 |
| (X = x) | 0 | -2 | 2 |
| P(Z = z) | 2/4 | 1/4 | 1/4 |
| P(Z ≤ z) | 2/4 | 2/4 + 1/4 = 3/4 | 2/4 + 1/4 + 1/4 = 1 |
Ques: Find the cumulative distribution function formula F(x), for the discrete random variable X.
Find: a) P(X = 3)
b) P(X > 2) (5 Marks)
| X | 1 | 2 | 3 | 4 | 5 |
| F(X) | 0.3 | 0.45 | 0.68 | 0.88 | 1 |
Ans. The Cumulative Distribution Function (CDF) of a random variable, X is given by
F(x) = P(X ≤ x)
- P(X = 3)
To get the CDF, we have to solve till we get the value less than or equal to 3.
From the table given above, we can get the value,
F(3) = P(X = 3) = P(X = 1) + P(X = 2) + P(X = 3)
From the table given above, the value of F(3) is 0.68.
- P(X > 2)
P(X > 2) = 1 - P(X ≤ 2)
P(X > 2) = 1 - F(2)
P(X > 2) = 1 - 0.45 [F(2) = 0.45]
P(X > 2) = 0.55
Ques: Consider Z be a discrete random variable with RZ = {1, 2, 3, 4….}. The PMF of Z is given by (5 Marks)
PZ(k) = 1/2k for k = 1,2,3,4….
Find P(2 < X ≤ 4)
Find P(X > 5)
k=1∑∞PX(k) = k=1∑∞ 1/2k = 1 (geometric mean)
Ans. In particular,
Therefore,
For x < 1, FX(x) = 0,
For 1 ≤ x ≤ 2, FX(x) = PX(1) = 1/2,
For 2 ≤ x < 3, FX(x) = PX(1) + PX(2) = 1/2 + 1/4 = 3/4
And so on,
In general, we have
For 0 < k ≤ x < k+1,
FX(x) = PX(1) + PX(2) + PX(3)+ …..+ PX(k)
= 1/2 + 1/4 + 1/8 +............1/2k = (2k-1)/2k
- To find P(2 < X ≤ 4), we have
P(2 < X ≤ 4) = FX(4) - FX(2) = 15/16 - 3/4 = 3/16.
Or we can write
P(2 < X ≤ 4) = PX(3) + PX(4) = 1/8 + 1/16 = 3/16
- To find Find P(X > 5), we can write
P(X > 5) = 1 - P(X ≤ 5) = 1 - FX(5) = 1 - 31/32 = 1/32
Ques: Find the Cumulative Distribution Function of the random variable
f(x) = k(x2 + 2x); if 0 ≤ x ≤ 1
0; else (5 Marks)
Ans. The random variable is given, To find the CDF we know,
-∞∫∞f(x)dx = 1….(1)
Therefore,
K0∫1 (x2 + 2x)dx = 1
K (x3/3 + X2)01 = 1
K(4/3) = 1
K = 3/4
The CDF can be integrated within the interval (-∞, x). If x lies in the interval (-∞, x) then,
F(x) = -∞∫x f(x)dx F(x)
F(x) = -∞∫x 0dx F(x) = 0
If x lies in the interval [0, 1] then,
F(x) = -∞∫x f(x)dx F(x) = -∞∫0 f(x)dx + 0∫x f(x)dx F(x) = 0 + 3/4(x3/3 + x2)
If x lies in the interval [1, ∞) then,
F(x) = -∞∫x f(x)dx F(x) = -∞∫0 f(x)dx + 0∫1 f(x)dx + 1∫x f(x)dx
F(x) = 0 + 3/4(x3/3 + x2)01 + 0F(x) = 3/4 x 4/3 = 1
Therefore, the CDF is given by
0; if x < 0
F(x) = 3/4(x3/3 + x2); if 0 ≤ x ≤ 1
1; if x > 1
Ques: The Cumulative Distribution Function of X is given by (5 Marks)
0; x < 1
F(x) = 1/4(x3 + 2x2 + 5x - a); 1 ≤ x ≤5
1; x>5
where a is constant.
Find the value of a.
Show that the lower quartile of X, qi, lies between 1.08 and 1.085. Here qi is a quarter of the probability.
Ans. CDF of variable X is given to us
0; x < 1
F(x) = 1/4(x3 + 2x2 + 5x - a); 1 ≤ x ≤5
1; x>5
(a) we know that,
F(x) = P(X ≤ x)....(1)
putting the lower limit ‘1’ in equation (1) we have,
F(1) = P(X ≤ 1)
F(1) = 0 [Probability below the lower limit is always 0]
Now, substitute F(1) = 0 in F(x); we have
1/4[(1)3 + 2(1)2 + 5(1) - a] = 0
1/4[8 - a] = 0\(\rightarrow\)a = 8
Thus, the value of a is 8.
(b) From the probability distribution function, we can say that the probability of being less than qi is 0.25 (1/4th of the total probability)
We can write as P(qi) = 0.25
Thus to prove F(1.08) < P(qi) < F(1.085)
Calculating,
F(1.08) = 1/4[(1.08)3 + 2(1.08)2 + 5(1.08) - 8]
F(1.08) = 0.247….(1)
Similarly,
F(1.085) = 1/4[(1.085)3 + 2(1.085) + 5(1.085) - 8]
F(1.085) = 0.253
Thus, we can conclude that
F(1.08) < F(qi) < F(1.085)
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