1. Probability Mass Function Expectation 郭俊利 2009/03/16
03. Random Variable
Probability Mass Function
Expectation
郭俊利 2009/03/16

2. Outline Review Random variable PMF Conditional PMF
Conditional probability
Random variable
PMF
Conditional PMF
1.7 ~ 2.6

3. Example 1
From the set of integers {1, 2, 3,. . . , } a number is selected at random. What is the probability that the sum of its digits is 11?
All – (a digit over 11) – (a digit over 10)
= H511 – C51 – C52 2!

4. Example 2
First throw an unbiased die, then throw as many unbiased coins as the point shown on the die.
What is the probability of obtaining k heads?
If 3 heads are obtained, what is the probability that the die showed n?

5. Random Variable
The random variable is a real-valued function of the outcome of the experiment.
 Discrete
 General = Continuous
p(x) 6 5 4 3 2 1
(a)
(b)
The sum of two dices is x,
what is p(x) ?
p(x) is uniform ? 36 x
… 7 …

6. Expected Value (1/2) Example:
Expectation E[X] = Σ x pX(x)
p(x) > 0
upper
upper
x ∈ X
lower
lower
Example:
The expectation of throwing a dice is 3.5
The answer to a question is 80% correctly, the grade may be 80.

7. Expected Value (2/2) E[a] = a E[aX] = aE[x] E[aX + b] = aE[x] + b
E[g(X)] = Σ g(x) px(x)
E[X] = Σ E[Xi] = np (p is uniform!)
i = 1 ~ n

8. Example 3 The shooting average of A is 2/3
The shooting average of B is 3/4
The shooting average of C is 4/5
(1) P (at least one hit) =
(2) P (one hit) =
P (two hits) =
P (three hits) =
(3) (A hit | one hit) =
(4) E (how many hits) =

9. { { PMF Probability Mass Function X = 1, if a head, 0, if a tail.
Its PMF is
pX(k) = p, if k = 1,
1 – p, if k = 0. {

10. Example 4
Let X be a random variable that takes values from 0 to 9 equal likely.
(1) Find the PMF of Y = X mod 3
(2) Find the PMF of Z = 5 mod X+1
A family has 5 natural children and has adopted 2 girls. Find the PMF of the number of girls out of the 7 children.

11. Example 5 (variance) { p(x) = x2 / a, if |x| < 4 and x ∈ Z
0, otherwise. {
(1) Find a and E[X].
(2) What is the PMF of Z = (X – E[X])2?
(3) From above, find the variance of X.
(4) var(X) = Σx (x – E[X])2 pX(x).

12. Important PMF Bernoulli Binomial Geometric Poisson pX(k) = p, 1-p
pX(k) = Cnk pk (1 – p)n – k
Geometric
pX(k) = pk (1 – p)n – k
Poisson
pX(k) = e–λλk / k!
e = …

13. Poisson Poisson is a good model for Binomial
When p is very very very small
and n is very very very large, then
λ = np
The probability of a wrong words is 0.1% and there are 1000 words in the document, what is the probability of 5 words found?
C (0.999)95 ≒ e–1 15 5!

14. Joint PMF Joint PMF Marginal PMF Example pX, Y(x, y) = P(X = x, Y = y)
pX(x) = Σy pX, Y(x, y)
pY(y) = Σx pX, Y(x, y)
Example
pX, Y(x, y) = 1 / 52
pX(x) = = 1 / 4 1 52

15. Example 6 X: –3 < x < 5 Question Y: –2 < y – x < 2
Joint PMF
pX, Y(x, y) =
Marginal PMF
pX(x) =
pY(y) =
The averages of X and Y
Question
X and Y ∈ Z

16. Example 7
A perfect coin is tossed n times. Let Yn denote the number of heads obtained minus the number of tails. Find the probability distribution of Yn and its mean.

17. Conditional PMF pX|A (x) = P(X = x | A) pX(x) = Σi P(Ai) pX|Ai (x)
pX, Y(x, y) = pY(y) pX|Y (x|y)
E[X] = Σi P(Ai) E[X | Ai]

18. Example 8
You maybe ask 0, 1 or 2 questions equal likely, I answer 75% correctly. X: the number of questions; Y: the number of wrong answers. Constructing joint PMF pX, Y(x, y) to find the probability of…
One question is asked and is answered wrong.
At least one wrong answer.