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45-733 PROBABILITY AND STATISTICS I Notes #2A

January 2000

Discrete Random Variables

1. Definition: Random Variable
A variable the value of which is a number determined by the outcome of an experiment.
2. Example: We flip a coin. Let X = 1 if Heads, and let X = 0 if Tails.
3. Example: We roll a die. Let X = "number of dots".
4. Definition: Probability Distribution: A function that assigns probabilities to the values of a random variable.
5. Example: We flip a coin. Given the definition of a random variable in (2), let the probability distribution be:
```

æ1/2 x = 0
ç
f(x) = ç1/2 x = 1
ç
è 0 otherwise
```
6. A plot of this function looks like this:

7. Note that it is essential that the function, f(x), be defined for all possible values of the argument, x -- in this case, the entire real line. Consequently, the "0 otherwise" is not trivial!
8. Example: We roll two dice. Let X = "sum of the two faces" and let the probability distribution be:
```
æ1/36  x = 2 or x = 12
ç
ç2/36  x = 3 or x = 11
ç
ç3/36  x = 4 or x = 10
ç
f(x) = ç4/36  x = 5 or x = 9
ç
ç5/36  x = 6 or x = 8
ç
ç6/36  x = 7
ç
è 0 otherwise
```
9. Properties of Discrete Probability Distributions
a) f(xi) ³ 0 for all i
b) åi=1,n f(xi) = 1
c) P[a £ X £ b] = å f(xi)
10. Example: With respect to (8):
P[6 £ X £ 9] = P[X = 6] + P[X = 7] + P[X = 8] + P[X = 9] =
1/36(5 + 6 + 5 + 4) = 20/36 = 5/9

Note that: P[6 < X £ 9] = P[7 £ X £ 9]

11. Definition: Distribution Function
F(x) = P(X £ x)
12. Example: With respect to (8):
F(5) = P[X = 1] + P[X = 2] + P[X = 3] + P[X = 4] + P[X = 5] =
1/36(1 + 2 + 3 + 4) = 10/36 = 5/18
13. Note that, when working with discrete distributions that:
P[X < 5] = F(4)
P[5 £ X £ 8] = F(8) - F(4)
P[5 £ X < 8] = F(7) - F(4)
P[5 < X < 8] = F(7) - F(5)
14. Discrete Uniform Distribution
```
æ1/n x = 1,2,3,4,5,...,n
f(x) = ç
è 0 otherwise
```
A graph of the function looks like this:

15. Examples of the Discrete Uniform Distribution
P[X £ 3] = 3/n
P[4 £ X £ n] = 1 - P[X < 4] = 1 - F(3) = (n-3)/n
F(5) = 5/n
16. Bernoulli Distribution
A Bernoulli experiment or trial is an experiment with only two possible outcomes. Traditionally these are known as a success and a failure. Let X = 1 be a success, and X = 0 be a failure, and let p be the probability: P(X = 1). Hence:
```
æ p  x = 1
ç
f(x) = ç1-p x = 0
ç
è 0 otherwise
```
17. Binomial Distribution
Let X1, X2, X3, ..., Xn be random variables corresponding to n Bernoulli trials (experiments), and let Z be the random variable:
Z = åi=1,n Xi
The distribution of Z is the binomial
```
æ ænö
ç ç ÷ pz(1-p)(n-z)
f(z) = ç èzø          z=0,1,2,3,...,n
ç
è 0 otherwise
```
18. Example: Let n = 4 and p = 1/2, then
P[Z = 2] = (4 choose 2)(1/2)2(1/2)2 = 6/16 = 3/8
19. Problem 3.1 p.79. Let A and B be the two impurities. We are given that P(A) = .4, P(B)= .5, and
1 - P(A È B) = .2. Hence
P(A È B) = .8 and
P(A È B) = P(A) + P(B) - P(A Ç B) = .4 + .5 - P(A Ç B) = .8. Hence
P(A Ç B) = .1
Let Y = number of impurites. Clearly P(Y = 2) = P(A Ç B) = .1
and P(Y = 0) = 1 - P(A È B) = .2. Hence
```
æ .2  y = 0
ç
ç .7  y = 1
f(y) = ç
ç .1  y = 2
ç
è  0 otherwise
```
To see that P(Y = 1) = .7, note that the shaded areas below correspond to this probability.

Hence, P(Y = 1) = P(A È B) - P(A Ç B) = .8 - .1 = .7
20. Problem 3.3 p.79
Let D = "Defective". Let Y be a random variable and let it equal the number of the test on which the 2nd defective is found. Clearly
P(Y = 2) = P(D on draw 1)* P(D on draw 2) = 1/2 * 1/3 = 1/6
P(Y = 3) = P(D on draw 1) * P(Dc on draw 2) * P(D on draw 3) + P(Dc on draw 2) * P(D on draw 1) * P(D on draw 3)
Hence: P(Y = 2) = 1/2*2/3*1/2 + 1/2*2/3*1/2 = 2/6
Using similar reasoning:
P(Y = 4) = 1/2*2/3*1/2*1 + 1/2*2/3*1/2*1 + 1/2*1/3*1*1 = 3/6
```
æ  1/6  y = 2
ç
ç  2/6  y = 3
f(y) = ç
ç  3/6  y = 4
ç
è   0 otherwise
```