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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.
Typology: Study notes
Uploaded on 04/15/2020
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Y 0 1 2 3 4 5 P(y) 1 6
5 18
2 9
1 6
1 9
1 18
๏ง Let ๐น ๐ฅ = ๐๐ฅ for ๐ฅ = 1, 2, 3, 4. Find ๐ so that ๐น(๐ฅ) satisfies the two properties of being a probability mass function. ๐น ๐ฅ ๐ฅโ๐
= ๐๐ฅ
4
๐ฅ=
= 1 1 = ๐ 1 + ๐ 2 + ๐ 3 + ๐ 4 1 = ๐ + 2๐ + 3๐ + 4๐ 1 = 10๐
1 10
๏งby substitution:
๐น ๐ฅ = 1 =
*therefore 0 โค ๐น ๐ฅ โค 1 (first property satisfied)
๐น ๐ฅ =
*therefore the sum of the probabilities is equal to 1 (second property satisfied)
๏งConsider the formula:
๐ ๐ฅ =
Show that ๐ ๐ฅ is indeed a pmf.
๏งConsider the formula:
๐ ๐ฅ =
Show that P ๐ฅ is indeed a pmf.
๏งalso referred to as the population mean ๏งdenoted by E ๐ฅ or ฮผ ๏งequal to the weighted average of the elements x in S where each element is weighted by its respective probability ๐ธ ๐ฅ = ๐ฅ ๐(๐ฅ) ๐ฅ๐๐
๏งLet x be a discrete rv and ๐ = *0, 1, 2, 3+. With the pmf of X given below,
P ๐ฅ =
1 4 , ๐๐ ๐ฅ ๐ ๐ 0, ๐๐ ๐ฅ โ 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 ๐ฅ =
1 6 ๐ฅ, ๐๐ ๐ฅ ๐ ๐ 0, ๐๐ ๐ฅ โ ๐ Compute the expected value of X
๐ธ ๐ฅ = ๐ฅ ๐ ๐ฅ ๐ฅ๐๐ = 1 16 1 + 2 16 2 + 3 16 3 =^16 +^46 +^96 =^146 ๐๐ 2^13