Applications of Probability?

83





A real estate agent is interested in the relationship between the number of lines in a newspaper advertisement for an apartment and the volume of inquiries from potential renters. Let volume of inquiries be denoted by the random variable X, with the value 0 for little interest, 1for moderate interest, and 2 for strong interest. The real estate agent estimated the joint probability function shown in the accompanying table.





Number of Number of Enquiries (X)





Lines (Y) 0 1 2


3 0.09 0.14 0.07


4 0.07 0.23 0.16


5 0.03 0.10 0.11





a. Find the joint cumulative probability function


b. Find and interpret the conditional probability function for Y, given X =0.


c. Find and interpret the conditional probability function for X, given Y= 5.


d. Find and interpret the covariance between X and Y.


e. Are number of lines in the advertisement and volume of inquiries independent of one another?

Applications of Probability?
a. Domain of X is (0, 1, 2) and domain of Y is (3, 4, 5).


Joint cumulative (or compound) distribution function is defined as:


F(y, x) = P[ Y %26lt;= y, X %26lt;= x]


In your matrix F(y, x) = sum of sub-matrix elements where sub-matrix consists of top left element until (y, x) element.


For instance F(4, 1) = 0.09 + 0.14 + 0.07 + 0.23 = 0.53.


b. P[Y = y|X = 0] is the first left column of your matrix.


c. P[X = x|Y = 5] is the last bottom row of your matrix.


d. This is toughy. Here it is:


P[X=0] = 0.19 = sum of the leftmost column. Similarly:


P[X=1] = 0.47


P[X=2] = 0.34..........(1)


Also:


P[Y=3] = 0.3 = sum of the top row. Similarly:


P[Y=4] = 0.46


P[Y=5] = 0.24..........(2)


Covariance is a measure of correlation of statistical variables. More positive =%26gt; variables are statistically similar meaning correlated (statistically dependent), more negative anti-correlated (again statistically dependent), and if zero they are not correlated (statistically independent).


Covariance is defined as:


Cov(X,Y) = sum(over X, Y)[(X - Mx)*(Y - My)*P(X,Y)]......(3), where Mx and My are mean values of X and Y respectively.


(3) can be rewritten as:


Cov(X,Y) = sum(over X,Y)[ X*Y*P(X,Y)] - Mx*My.......(4)


From (1) we get Mx = 1*0.47 + 2*0.34 = 1.15.......(5).


From (2) we get My = 3*0.3 + 4*0.46+ 5*0.24 = 3.94.....(6).


Plugging (5) and (6) into (4) (using your matrix) we get:


Cov(X,Y) = 0.109,


e. which shows that X and Y are pretty statistically independent, meaning that tenant interest does not depend much of the number of lines.
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