Probability Distributions and Expected Values of Discrete Random Variables, Study notes of Mathematics

This document illustrates the concept of random variables, both discrete and continuous, and their probability distributions. It also discusses the construction of probability mass functions and the calculation of probabilities for discrete random variables. The document then provides examples of calculating the expected values of discrete random variables.

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๏‚งillustrate a random variable (discrete and
continuous)
๏‚งdistinguish between a discrete and a
continuous random variable
๏‚งfind possible values of a random variable
๏‚งillustrate probability distributions for discrete
random variable and its properties
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๏‚งillustrate a random variable (discrete and

continuous)

๏‚งdistinguish between a discrete and a

continuous random variable

๏‚งfind possible values of a random variable

๏‚งillustrate probability distributions for discrete

random variable and its properties

An experiment consists of rolling a pair of dice.

Let Y be the random variable representing the

absolute value of the difference of the numbers

that come up. Construct the probability

distribution of Y in tabular, graphical and

formula form.

๏‚งthe values of random variables for which the

probability is greater than zero: ๐‘ƒ ๐‘ฅ > 0

*the mass points are: 0, 1, 2, 3, 4 and 5

Y 0 1 2 3 4 5 P(y) 1 6

5 18

2 9

1 6

1 9

1 18

The possible values (Support S) of a random

variable is the set of values that a random

variable can take. This means that the elements

of S can be put into one-to-one correspondence

with the set of natural numbers.

๏‚ง Let ๐น ๐‘ฅ = ๐‘˜๐‘ฅ for ๐‘ฅ = 1, 2, 3, 4. Find ๐‘˜ so that ๐น(๐‘ฅ) satisfies the two properties of being a probability mass function. ๐น ๐‘ฅ ๐‘ฅโˆˆ๐‘†

= ๐‘˜๐‘ฅ

4

๐‘ฅ=

= 1 1 = ๐‘˜ 1 + ๐‘˜ 2 + ๐‘˜ 3 + ๐‘˜ 4 1 = ๐‘˜ + 2๐‘˜ + 3๐‘˜ + 4๐‘˜ 1 = 10๐‘˜

1 10

๏‚งby substitution:

๐น ๐‘ฅ = 1 =

10 ๐น ๐‘ฅ = 3^ =

10 ๐น ๐‘ฅ = 4^ =

*therefore 0 โ‰ค ๐น ๐‘ฅ โ‰ค 1 (first property satisfied)

๐น ๐‘ฅ =

*therefore the sum of the probabilities is equal to 1 (second property satisfied)

๏‚งConsider the formula:

๐‘ƒ ๐‘ฅ =

Show that ๐‘ƒ ๐‘ฅ is indeed a pmf.

๏‚งConsider the formula:

๐‘ƒ ๐‘ฅ =

Show that P ๐‘ฅ is indeed a pmf.

๏‚งalso referred to as the population mean ๏‚งdenoted by E ๐‘ฅ or ฮผ ๏‚งequal to the weighted average of the elements x in S where each element is weighted by its respective probability ๐ธ ๐‘ฅ = ๐‘ฅ ๐‘(๐‘ฅ) ๐‘ฅ๐œ–๐‘†

๏‚งLet x be a discrete rv and ๐‘† = *0, 1, 2, 3+. With the pmf of X given below,

P ๐‘ฅ =

1 4 , ๐‘–๐‘“ ๐‘ฅ ๐œ– ๐‘† 0, ๐‘–๐‘“ ๐‘ฅ โˆ‰ ๐‘† Compute the expected value of X

๏‚งLet x be a discrete rv and ๐‘† = *1, 2, 3+. With the pmf of X given below,

P ๐‘ฅ =

1 6 ๐‘ฅ, ๐‘–๐‘“ ๐‘ฅ ๐œ– ๐‘† 0, ๐‘–๐‘“ ๐‘ฅ โˆ‰ ๐‘† Compute the expected value of X

๐ธ ๐‘ฅ = ๐‘ฅ ๐‘ ๐‘ฅ ๐‘ฅ๐œ–๐‘† = 1 16 1 + 2 16 2 + 3 16 3 =^16 +^46 +^96 =^146 ๐‘œ๐‘Ÿ 2^13