Random Variables
Discrete Probability Distributions
Continuous Random Variables
Lecture 3: Probability Distributions and Probability Densities - 1 Assist. Prof. Dr. Emel YAVUZ DUMAN MCB1007 Introduction to Probability and Statistics ˙Istanbul K¨ ult¨ ur University
Random Variables
Discrete Probability Distributions
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Discrete Probability Distributions
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Continuous Random Variables
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Outline
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Random Variables
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Discrete Probability Distributions
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Continuous Random Variables
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 1 If S is a sample space with a probability measure and X is a real-valued function defined over the elements of S, then X is called a random variable (or stochastic variable). In this course we shall always denote random variables by capital letters such as X , Y etc., and their values by the corresponding lowercase letters such as x and y , respectively.
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Example 2 Suppose that a coin is tossed twice so that the sample space is S = {HH, HT , TH, TT }. Let X represent the number of heads that can come up. With each sample point we can associate a number for X as shown in the table: Sample Point Probability x
HH
HT
TH
TT
1 4
1 4
1 4
1 4
2
1
1
0
Thus, for example, in the case of HH (i.e., 2 heads), X = 2 while for TH (1 head), X = 1. It follows that X is a random variable. Also, we can write P(X = 2) = 14 , P(X = 1) = 14 + 14 = 12 , and P(X = 0) = 14 .
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Discrete Probability Distributions
Continuous Random Variables
Example 3 A balanced coin is tossed four times. List the elements of the sample space that are presumed to be equally likely, as this is what we mean by a coin being balanced, and the corresponding values x of the random variable X , the total number of heads. Solution. If H and T stand for heads and tails, the results are as shown in the following table:
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Elements of sample space HHHH HHHT HHTH HTHH THHH HHTT HTHT HTTH
Discrete Probability Distributions
Probability 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16
x 4 3 3 3 3 2 2 2
Elements of sample space THHT THTH TTHH HTTT THTT TTHT TTTH TTTT
Continuous Random Variables
Probability
1 4 Thus, we can write P(X = 0) = 16 , P(X = 1) = 16 , 6 4 1 . P(X = 2) = 16 , P(X = 3) = 16 and P(X = 4) = 16
1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16
x 2 2 2 1 1 1 1 0
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Discrete Probability Distributions
Continuous Random Variables
Example 4 Two socks are selected at random and removed in succession from a drawer containing five brown socks and three green socks. List the elements of the sample space, the corresponding probabilities, and the corresponding values x of the random variable X is the number of brown socks selected. Solution. If B and G stand for brown and green, then we have following probabilities 20 5 3 15 5 4 P(BB) = · = , P(BG ) = · = , 8 7 56 8 7 56 15 3 2 6 3 5 P(GB) = · = , and P(GG ) = · = , 8 7 56 8 7 56 and the results are shown in the following table: Elements of Sample Space Probability x
BB 20/56 2
BG 15/56 1
GB 15/56 1
GG 6/56 0
Random Variables
Discrete Probability Distributions
Continuous Random Variables
In all of the examples of this section we have limited our discussion to discrete sample space, and hence to discrete random variable, namely, random variables whose range is finite or countably infinite. Continuous random variables defined over continuous sample spaces will be taken up in the third section.
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Continuous Random Variables
Random Variables
Discrete Probability Distributions
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Definition 5 If X is a discrete random variable, the function given by f (x) = P(X = x) for each x within the range of X is called the probability distribution (or probability function) of X . Based on the postulates of probability, it immediately follows that Theorem 6 A function can serve as the probability distribution of a discrete random variable X if and only if its values, f (x), satisfy the conditions 1 2
f (x) ≥ 0 for each value within its domains; x f (x) = 1, where the summation extends over all the values within its domain.
Random Variables
Discrete Probability Distributions
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Example 7 Find a formula for the probability distribution of the total number of heads obtained in four tosses of a balanced coin. 1 4 , P(X = 1) = 16 , Solution. We know that P(X = 0) = 16 6 4 1 P(X = 2) = 16 , P(X = 3) = 16 and P(X = 4) = 16 . Observing that the numerators of these 4 fractions, 4 4 1, 4,46, 4, and 1, are 4 five the binomial coefficients 0 , 1 , 2 , 3 , and 4 , we find that the formula for the probability distribution can be written as 4
f (x) =
x
16
for x = 0, 1, 2, 3, 4.
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Discrete Probability Distributions
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Example 8 Check whether the function given by f (x) =
x +2 for x = 1, 2, 3, 4, 5 25
can serve as the probability distribution of a discrete random variable. Solution. Substituting the different values of x, we get f (1) = 4 5 6 7 , f (3) = 25 , f (4) = 25 , and f (5) = 25 . Since these f (2) = 25 values are all nonnegative, the first condition of Theorem 6 is satisfied, and since f (1) + f (2) + f (3) + f (4) + f (5) =
3 25 ,
4 5 6 7 3 + + + + =1 25 25 25 25 25
the second conditions of Theorem 6 is satisfied. Thus, the given function can serve as the probability distribution of a random variable having the range {1, 2, 3, 4, 5}.
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Example 9 Suppose that a pair of fair dice are to be tossed, and let the random variable X denote the sum of the points. Obtain the probability distribution for X . Solution. The random variable X is the sum of the coordinates for each point. Thus for (3, 2) we have X = 5. Using the fact that all 36 sample points are equally probable, so that each sample point has probability 1/36. First Die
Second Die
The one is red is the sum
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x f (x)
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Continuous Random Variables
2
3
4
5
6
7
8
9
10
11
12
1 36
2 36
3 36
4 36
5 36
6 36
5 36
4 36
3 36
2 36
1 36
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Figure 1 : Probability Bar Chart
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 10 If X is a discrete random variable, the function given by f (t) for − ∞ < x < ∞ F (x) = P(X ≤ x) = t≤x
where f (t) is the value of the probability distribution of X at t, is called the distribution function, or the cumulative distribution, of X.
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Discrete Probability Distributions
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Based on the postulates of probability and some of their immediate consequences, it follows that Theorem 11 The values F (x) of the distribution function of a discrete random variable X satisfy the conditions 1
F (−∞) = 0 and F (∞) = 1;
2
if a < b, then F (a) ≤ F (b) for any real numbers a and b.
If we are given the probability distribution of a discrete random variable, the corresponding distribution function is generally easy to find.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 12 Find the distribution function of the total of heads obtained in four tosses of a balanced coin. 1 4 , f (1) = 16 , f (2) = Solution. Given f (0) = 16 1 f (4) = 16 from Example 3, it follows that
F (0) = f (0) =
6 16 ,
f (3) =
1 , 16
F (1) = f (0) + f (1) =
5 , 16
F (2) = f (0) + f (1) + f (2) =
11 , 16
15 , 16 F (4) = f (0) + f (1) + f (2) + f (3) + f (4) = 1. F (3) = f (0) + f (1) + f (2) + f (3) =
4 16 ,
and
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Discrete Probability Distributions
Hence, the distribution function ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪1 ⎪ 16 ⎪ ⎪ ⎨5 F (x) = 16 11 ⎪ ⎪ 16 ⎪ ⎪ ⎪ 15 ⎪ ⎪ ⎪ 16 ⎪ ⎩1
Continuous Random Variables
is given by for for for for for for
x < 0, 0 ≤ x < 1, 1 ≤ x < 2, 2 ≤ x < 3, 3 ≤ x < 4, x ≥ 4.
Observe that this distribution function is defined not only for the values taken on by the given random variable, but for all real 5 and F (100) = 1, numbers. For instance, we can write F (1.7) = 16 although the probabilities of getting at most 1.7 heads or at most 100 heads in four tosses of a balanced coin may not be of any real significance.
Random Variables
Discrete Probability Distributions
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Example 13 Find the distribution function of the random variable X of Example 4 and plot its graph. Solution. Based on the probabilities given in the following table Elements of Sample Space Probability x we can write f (0) = that
6 56 ,
f (1) =
F (0) = f (0) =
BB 20/56 2 15 56
+
15 56
BG 15/56 1 =
30 56 ,
GB 15/56 1
GG 6/56 0
and f (2) =
20 56 ,
6 , 56
36 , 56 F (2) = f (0) + f (1) + f (2) = 1. F (1) = f (0) + f (1) =
so
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Discrete Probability Distributions
Hence, the distribution function ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨6 F (x) = 56 36 ⎪ ⎪ ⎪ 56 ⎪ ⎩1
Continuous Random Variables
of X is given by for for for for
x < 0, 0 ≤ x < 1, 1 ≤ x < 2, x ≥ 2.
F (x) 1
36/56
6/56 0
x 1
2
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Discrete Probability Distributions
Continuous Random Variables
Example 14 Find the distribution function of the random variable that has the probability distribution f (x) = Solution. Since f (1) = 5 , then f (5) = 15
x for x = 1, 2, 3, 4, 5. 15
1 15 ,
F (x) =
f (2) =
⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ 15 ⎪ ⎪ ⎨3 15
6 ⎪ ⎪ 15 ⎪ ⎪ ⎪ 10 ⎪ ⎪ 15 ⎪ ⎪ ⎩1
2 15 ,
for for for for for for
f (3) =
3 15 ,
x < 1, 1 ≤ x < 2, 2 ≤ x < 3, 3 ≤ x < 4, 4 ≤ x < 5, x ≥5
f (4) =
4 15 ,
and
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Theorem 15 If the range of a random variable X consists of the values x1 < x2 < x3 < · · · < xn , then f (x1 ) = F (x1 ) and f (xi ) = F (xi ) − F (xi −1 ) for i = 2, 3, · · · , n.
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Continuous Random Variables
Example 16 If X has the distribution function F (1) = 0.25, F (2) = 0.61, F (3) = 0.83, and F (4) = 1 for x = 1, 2, 3, 4, find the probability distribution of X . Solution. We have f (1) = F (1) = 0.25, f (2) = F (2) − F (1) = 0.61 − 0.25 = 0.36, f (3) = F (3) − F (2) = 0.83 − 0.61 = 0.22, f (4) = F (4) − F (3) = 1 − 0.83 = 0.17.
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Discrete Probability Distributions
Example 17 If X has the distribution function ⎧ ⎪ 0 for ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎨ 4 for F (x) = 12 for ⎪ ⎪ ⎪ 3 for ⎪ ⎪ 4 ⎪ ⎩ 1 for
x < −1, − 1 ≤ x < 1, 1 ≤ x < 3, 3 ≤ x < 5, x ≥ 5.
find 1
P(X ≤ 3), P(X = 3), P(X < 3);
2
P(X ≥ 1);
3
P(−0.4 < X < 4);
4
P(X = 5);
5
the probability distribution of X .
Continuous Random Variables
Random Variables
Discrete Probability Distributions
⎧ ⎪ 0 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎨4 F (x) =
1
2 ⎪ ⎪ 3 ⎪ ⎪ ⎪ 4 ⎪ ⎩ 1
for for for for for
x < −1, − 1 ≤ x < 1, 1 ≤ x < 3, 3 ≤ x < 5, x ≥ 5.
Solution. 1
3 4 1 3 1 P(X = 3) = − = 4 2 4 1 P(X < 3) = 2 P(X ≤ 3) =
Continuous Random Variables
Random Variables
Discrete Probability Distributions
⎧ ⎪ 0 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎨4 F (x) =
1
2 ⎪ ⎪ 3 ⎪ ⎪ ⎪ 4 ⎪ ⎩ 1
for for for for for
x < −1, − 1 ≤ x < 1, 1 ≤ x < 3, 3 ≤ x < 5, x ≥ 5.
2
P(X ≥ 1) = 1 − P(X < 1) = 1 −
3
P(−0.4 < X < 4) =
4 5
Continuous Random Variables
1 4
3 1 1 4 − 4 = 2. P(X = 5) = 1 − 34 = 14 . f (−1) = 14 , f (1) = 12 − 14 = 14 , f (3) f (5) = 1 − 34 = 14 , and 0 elsewhere.
= 34 .
=
3 4
−
1 2
= 14 ,
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Continuous Random Variables
Random Variables
Discrete Probability Distributions
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Continuous Random Variables
So far we have considered discrete random variables that can take on a finite or countably infinite number of values. In applications, we are often interested in random variables that can take on an uncountable continuum of values; we call these continuous random variables. Consider modeling the distribution of the age that a person dies at. Age of death, measured perfectly with all the decimals and no rounding, is a continuous random variable (e.g., age of death could be 87.3248583585642 years). Other examples of continuous random variables include: time until the occurrence of the next earthquake in ˙Istanbul; the lifetime of a battery; the annual rainfall in Ankara.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Because it can take on so many different values, each value of a continuous random variable winds up having probability zero. If I ask you to guess someone’s age of death perfectly, not approximately to the nearest millionth year, but rather exactly to all the decimals, there is no way to guess correctly - each value with all decimals has probability zero. But for an interval, say the nearest half year, there is a nonzero chance you can guess correctly. So, we have the following definition: Definition 18 A random variable X is called a continuous random variable if its distribution function F is a continuous function on R, or equivalently, if P(X = x) = 0, for every x ∈ R.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
The definition of probability in the continuous case presumes for each random variable the existence of a function, called a probability density function (pdf), such that areas under the curve give the probabilities associated with the corresponding intervals along the horizontal axis. In other words, a probability density function, integrated form a to b (with a ≤ b), gives the probability that the corresponding random variable will take on a value on the interval from a to b. Definition 19 A function with values f (x), defined over the set of all real numbers, is called a probability density function (pdf) of the continuous random variable X if and only if
b
P(a ≤ X ≤ b) =
f (x)dx a
for any real constants a and b with a ≤ b.
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Discrete Probability Distributions
Continuous Random Variables
f (x)
a
b
x
Figure 2 : P(a ≤ X ≤ b) = area under the density curve between a and b.
Theorem 20 If X is a continuous random variable and a and b are real constants with a ≤ b, then P(a ≤ X ≤ b) = P(a ≤ X < b) = P(a < X ≤ b) = P(a < X < b).
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Discrete Probability Distributions
Continuous Random Variables
Analogous to Theorem 6, let us now state the following properties of probability densities, which again follow directly from the postulates of probability. Theorem 21 A function can serve as a probability density of a continuous random variable X if its values, f (x), satisfy the conditions 1 2
f (x) ≥ 0 for −∞ < x < ∞, ∞ −∞ f (x)dx = 1.
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Discrete Probability Distributions
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Example 22 The probability density of the continuous random variable X is given by 1 for 2 < x < 7, f (x) = 5 0 elsewhere. 1
2
Draw its graph and verify that the total area under the curve (above the x-axis) is equal to 1. Find P(3 < X < 5).
Solution. f (x) 1 5
2
7
x
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Discrete Probability Distributions
f (x) =
1
1 5
0
Continuous Random Variables
for 2 < x < 7, elsewhere.
It is clear that f (x) ≥ 0 for all x and
∞
f (x)dx = −∞
2
f (x) dx + −∞
2
0
7
= 2
7
f (x) dx +
1 5
∞ 7
1 7 1 1 dx = x = (7 − 2) = 1. 5 5 2 5
f (x) dx
0
2
5
P(3 < X < 5) =
f (x)dx = 3
2 = . 5
3
5
1 5 1 1 dx = x = (5 − 3) 5 5 3 5
Random Variables
Discrete Probability Distributions
Example 23 The pdf of the random variable X is given by √c for 0 < x < 4, x f (x) = 0 elsewhere. Find 1
the value of c,
2
P(X < 0.25) and P(X > 1).
Continuous Random Variables
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Discrete Probability Distributions
f (x) =
√c x
for 0 < x < 4,
0
elsewhere.
Continuous Random Variables
Solution. 1
To satisfy the second condition of Theorem 21, we must have
∞
f (x)dx = −∞
0
f (x) dx + −∞ 0
0
4
f (x) dx +
√c
x
∞ 4
f (x) dx
0
4
4 − 12 +1 1 x c √ dx = cx − 2 dx = c 1 = x −2 + 1 0 0 0 √ = 2c x|40 = 4c = 1
4
and it follows that c = 1/4.
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Discrete Probability Distributions
f (x) =
1 √ 4 x
for 0 < x < 4,
0
elsewhere.
Continuous Random Variables
2
P(X < 0.25) =
0.25
f (x)dx = −∞
0.25
= 0
1 = . 4
0
f (x) dx + −∞ 0
0
0.25
f (x) dx
1 √ 4 x
1 √ 0.25 1 √ 1 √ dx = 2 x = ( 0.25 − 0) 4 2 4 x 0
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Discrete Probability Distributions
f (x) =
1 √ 4 x
for 0 < x < 4,
0
elsewhere.
Continuous Random Variables
2
P(X < 1) =
∞
f (x)dx = 1
= 1
1 = . 2
1 4
4
f (x) dx +
1 √ 4 x
4
∞
f (x) dx
1 √ 4 1 1 √ dx = 2 x = (2 − 1) 4 2 4 x 1
0
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Discrete Probability Distributions
Continuous Random Variables
Example 24 If X has the probability density ke −3x f (x) = 0
for x > 0, elsewhere,
find k and P(0.5 ≤ X ≤ 1). Solution. To satisfy the second condition of Theorem 21, we must have
t e −3x f (x)dx = ke dx = k lim t→∞ −3 −∞ 0 0 k k k = − lim (e −3t − e 0 ) = − (0 − 1) = = 1 t→∞ 3 3 3 ∞
∞
and it follows that k = 3.
−3x
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3e −3x f (x) = 0
Discrete Probability Distributions
Continuous Random Variables
for x > 0, elsewhere,
For the probability we get
P(0.5 ≤ X ≤ 1) =
1
3e 0.5
−3x
1 e −3x dx = 3 = −e −3 + e −1.5 −3 0.5
≈ 0.1733430918.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 25 If X is a continuous random variable and the value of its probability density at t is f (t), then the function given by
x f (t)dt for − ∞ < x < ∞ F (x) = P(X ≤ x) = −∞
is called the distribution function, or the cumulative distribution, of X . The values F (x) of the distribution function of a continuous random variable X satisfies the conditions: F (−∞) = 0, F (∞) = 1, and F (a) ≤ F (b) when a < b.
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Theorem 26 If f (x) and F (x) are the values of the probability density and the distribution function of X at x, then P(a ≤ X ≤ b) = F (b) − F (a) for any real constants a and b with a ≤ b, and f (x) = where the derivative exists.
dF (x) dx
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Discrete Probability Distributions
Continuous Random Variables
Example 27 Find the distribution function of the random variable X of Example 24, and use it to reevaluate P(0.5 ≤ x ≤ 1). Solution. For x > 0,
x f (t)dt = F (x) = −∞
0
x
x 3e −3t dt = −e −3t 0 = 1 − e −3x
and since F (x) = 0 for x ≤ 0, we can write 0 for x ≤ 0, F (x) = −3x for x > 0. 1−e
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Discrete Probability Distributions
0 F (x) = 1 − e −3x
Continuous Random Variables
for x ≤ 0, for x > 0.
To determine the probability P(0.5 ≤ X ≤ 1), we use the first part of Theorem 26, getting P(0.5 ≤ X ≤ 1) = F (1) − F (0.5) = (1 − e −3 ) − (1 − e −1.5 ) = −e −3 + e −1.5 ≈ 0.1733430918. This agrees with the result obtained by using the probability density directly in Example 24.
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Discrete Probability Distributions
Continuous Random Variables
Example 28 Find the distribution function of the random variable X whose probability density is given by ⎧ ⎪ for 0 < x < 1, ⎨x f (x) = 2 − x for 1 ≤ x < 2, ⎪ ⎩ 0 elsewhere.
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Discrete Probability Distributions
⎧ ⎪ ⎨x f (x) = 2 − x ⎪ ⎩ 0
for 0 < x < 1, for 1 ≤ x < 2, elsewhere.
Solution. For 0 < x < 1,
x f (t)dt = F (x) = −∞
x
0
for 1 ≤ x < 2,
x f (t)dt = F (x) = −∞ 2 1
Continuous Random Variables
x t 2 x2 tdt = = , 2 0 2
1
f (t)dt +
0
x
f (t)dt = 1
1
x
tdt + 0
1
(2 − t)dt
x x2 1 x2 t2 1 t − 2 + = − + 2x − 1, = + 2x − = + 2t − 2 0 2 1 2 2 2 2 and for F (x) = 1 where x ≥ 2.
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Discrete Probability Distributions
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So, we have ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨ x2 F (x) =
2
⎪ − x2 + 2x − 1 ⎪ ⎪ ⎪ ⎩1 2
for for for for
x ≤ 0, 0 < x < 1, 1 ≤ x < 2, x ≥ 2.
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Discrete Probability Distributions
Continuous Random Variables
Example 29 Find a probability density function for the random variable whose distribution function is given by ⎧ ⎪ ⎨0 for x ≤ 0, F (x) = x for 0 < x < 1, ⎪ ⎩ 1 for x ≥ 1. Solution. Since the given density function is differentiable everywhere except at x = 0 and x = 1, we differentiate for x < 0, 0 < x < 1, and x > 1, getting 0, 1, and 0. Thus, we can write ⎧ ⎪ ⎨0 for x ≤ 0, f (x) = 1 for 0 < x < 1, ⎪ ⎩ 0 for x ≥ 1.
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To fill the gaps at x = 0 and x = 1, we let f (0) and f (1) both equal to zero. Actually, it does not matter how the probability density is defined at these points, but there are certain advantages for choosing the values in such a way that the probability density is nonzero over an open interval. Thus, we can write the probability density of the original random variable as 1 for 0 < x < 1, f (x) = 0 elsewhere.
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Discrete Probability Distributions
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Example 30 The distribution function of the random variable Y is given by 1 − y92 for y > 3, F (y ) = 0 elsewhere. Find P(Y ≤ 5), P(Y > 8) and a probability density function of Y . Solution. P(Y ≤ 5) = F (5) = 1 −
9 52
=
16 25 ,
P(Y > 8) = 1 − P(Y ≤ 8) = 1 − F (8) = 1 − 1 + d 9 for y > 3, 1 − = 18 y2 y3 f (y ) = dy 0 elsewhere.
9 64
=
9 64 ,
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Discrete Probability Distributions
Example 31 The distribution function of the ⎧ ⎪ ⎨0 F (x) = x+1 2 ⎪ ⎩ 1
Continuous Random Variables
random variable X is given by for x < −1, for − 1 < x < 1, for x ≥ 1.
Find P − 12 < X < 12 and P(2 < X < 3). Solution. P − 12 < X < 12 = F 12 − F − 12 = 3 1 1 4 − 4 = 2.
1 +1 2
2
P(2 < X < 3) = F (3) − F (2) = 1 − 1 = 0.
−
− 12 +1 2
=
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Discrete Probability Distributions
Thank You!!!
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