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Marginal Distribution and Beta-Binomial: Getting City Ranking Probabilities and Moments - , Study notes of Economic statistics

University of Illinois - Urbana-ChampaignEconomic statistics
Prof. Vidisha N. Vachharajani

An example of calculating marginal distributions from joint distributions using the national opinion research center amalgam survey of 1972 data. The example focuses on city rankings and introduces the concept of the beta-binomial distribution. The document also covers the properties of conditional expectation and conditional variance for the beta-binomial distribution.

Typology: Study notes

Pre 2010

Uploaded on 03/16/2009

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TA Session 4
ECON 506
TA: Vidisha Vachharajani
October 2, 2008
1 Marginal Distribution from Joint
We’ll consider another useful example of getting the marginal distribution from the given joint distribution.
Example 1 (Ranks) The National Opinion Research Center Amalgam Survey of 1972 asked people to rank
three kinds of areas - City (over 50,000), Suburb (within 30 miles of a city), and Country (everywhere else).
The results are below, with respondents categorized as living in their respective current residence.
Ranks Residence
(City, Suburb, Country) City Suburb Country Total
(1,2,3) 210 22 10 242
(1,3,2) 23 4 1 28
(2,1,3) 111 45 14 170
(2,3,1) 8 4 0 12
(3,1,2) 204 299 125 628
(3,2,1) 81 126 152 359
Total 637 500 302 1439
Ranking of (1,2,3) implies that the respondent ranks city as best, then suburb and last country. Such a
ranking was given by 242 people, 210 of whom live in the city, 22 in the suburb and 10 in the country. Thus,
our random vector is (X, Y, Z ), for Xbeing City rank, Ybeing Suburb rank, and Zbeing Country rank. The
joint space for this vector, R
3
, comprises of the six permutations of 1,2and 3, and is seen in column 1in
the above table. Then we’re given that our of our population of 1439, if we at random choose one person,
the joint distribution of this individual’s ranking is:
f(x, y, z) = P[(X, Y , Z) = (x, y, z)] (1)
= 242/1439 if (x, y, z) = (1,2,3) (2)
= 28/1439 if (x, y, z) = (1,3,2) (3)
= 170/1439 if (x, y, z) = (2,1,3) (4)
= 12/1439 if (x, y, z) = (2,3,1) (5)
= 628/1439 if (x, y, z) = (3,1,2) (6)
= 359/1439 if (x, y, z) = (3,2,1) (7)
Then, if we want marginals of only the city ranking X, we can add up the joints of the rankings across
Yand Z, so that for each ranking 1,2and 3, we have:
f
X
(1) = f(1,2,3) + f(1,3,2) = 28 + 242
1439 = 0.1876 (8)
f
X
(2) = f(2,1,3) + f(2,3,1) = 170 + 12
1439 = 0.1265 (9)
f
X
(3) = f(3,1,2) + f(3,2,1) = 628 + 359
1439 = 0.6859 (10)
1
pf2

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Download Marginal Distribution and Beta-Binomial: Getting City Ranking Probabilities and Moments - and more Study notes Economic statistics in PDF only on Docsity!

TA Session 4

ECON 506

TA: Vidisha Vachharajani

October 2, 2008

1 Marginal Distribution from Joint

We’ll consider another useful example of getting the marginal distribution from the given joint distribution.

Example 1 (Ranks) The National Opinion Research Center Amalgam Survey of 1972 asked people to rank three kinds of areas - City (over 50,000), Suburb (within 30 miles of a city), and Country (everywhere else). The results are below, with respondents categorized as living in their respective current residence. Ranks Residence (City, Suburb, Country) City Suburb Country Total (1, 2 , 3) 210 22 10 242 (1, 3 , 2) 23 4 1 28 (2, 1 , 3) 111 45 14 170 (2, 3 , 1) 8 4 0 12 (3, 1 , 2) 204 299 125 628 (3, 2 , 1) 81 126 152 359 Total 637 500 302 1439 Ranking of (1, 2 , 3) implies that the respondent ranks city as best, then suburb and last country. Such a ranking was given by 242 people, 210 of whom live in the city, 22 in the suburb and 10 in the country. Thus, our random vector is (X, Y, Z), for X being City rank, Y being Suburb rank, and Z being Country rank. The joint space for this vector, R^3 , comprises of the six permutations of 1 , 2 and 3 , and is seen in column 1 in the above table. Then we’re given that our of our population of 1439 , if we at random choose one person, the joint distribution of this individual’s ranking is:

f (x, y, z) = P [(X, Y, Z) = (x, y, z)] (1) = 242 / 1439 if (x, y, z) = (1, 2 , 3) (2) = 28 / 1439 if (x, y, z) = (1, 3 , 2) (3) = 170 / 1439 if (x, y, z) = (2, 1 , 3) (4) = 12 / 1439 if (x, y, z) = (2, 3 , 1) (5) = 628 / 1439 if (x, y, z) = (3, 1 , 2) (6) = 359 / 1439 if (x, y, z) = (3, 2 , 1) (7)

Then, if we want marginals of only the city ranking X, we can add up the joints of the rankings across Y and Z, so that for each ranking 1 , 2 and 3 , we have:

fX (1) = f(1, 2 , 3) + f(1, 3 , 2) =

fX (2) = f(2, 1 , 3) + f(2, 3 , 1) =

fX (3) = f(3, 1 , 2) + f(3, 2 , 1) =

Thus, looking at the above marginal distribution of city rankings, what can we conclude? For how much of time is the City ranked third?

2 Beta-Binomial

We will use this distribution to illustrate the properties of conditional expectation and conditional variance, vis-a-vis:

E[Y ] = EX EY (Y |X) (11)

V ar(Y ) = EX (V arY (Y |X)) + V arX (EY (Y |X)). (12)

Consider two random variables, X and Y. Suppose (Y | X = x) ∼ Binomial(n, x) and X ∼ Beta(α, β). Then the marginal distribution of Y is the beta-binomial, so that:

Y ∼ Beta − Binomial(α, β, n). (13)

We already know the form of the binomial density function. The beta for X (marginal) has the following form:

fX (x) =

Γ(α + β) Γ(α)Γ(β) xα−^1 (1 − x)β−^1 , 0 < x < 1. (14)

Finally, the marginal for Y following the beta-binomial is:

fY (y) =

n y

Γ(y + α)Γ(n − y + β) Γ(n + α + β)

, y = 0, 1 , ..., n. (15)

Now since Y is conditionally binomial, we have:

EY (Y |X) = nx; V arY (Y |X) = nx(1 − x) (16)

Also, for X following the beta distribution unconditionally, we have:

E[X] = α α + β

For Y following beta-binomial as the marginal, moments of the beta-binomial are given by:

E[Y ] =

nα α + β

; V ar(Y ) =

nαβ(n + α + β) (α + β)^2 (1 + α + β)

You now have all the pieces necessary to verify the above shown properties (as in (11) and (12)) of conditional expectation and conditional variance! We already went over the variance part in class - just some algebra.