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Economics 671 Solutions: Problem Set # Multivariate and Conditional Probability
(1) (a) First, note that the density is always positive over its support. Moreover, ∫ (^) ∞ 0
x
2 exp[−(x + y)]dydx = 1
(which you can verify through some rather simple integration). Thus, this is indeed a valid density function.
(b) Note:
fY (y) =
∫ (^) y 0
2 exp[−(x + y)]dx = 2 exp(−y)[1 − exp(−y)].
Simiarly, fX (x) =
x
2 exp[−(x + y)]dy = 2 exp(− 2 x).
(c) The variables are not independent; the product of the marginal densities does not equal the joint density. That the variables must be dependent is rather obvious since the (conditional) support of X depends on Y.
(2) No, in general, zero correlation does not imply independence, though this is true for normal random variables, (and the converse of this statement is always true). To see this, consider two random variables: X ∼ N (0, 1) and Y = X^2 ∼ χ^2 (1) (That is, Y has a Chi-square distribution with one degree of freedom). Note that Cov(X, Y ) = E(XY ) − E(X)E(Y ) = E(XY ) = E(X^3 ) = 0, since all odd-order moments of the normal density is zero. So, Corr(X, Y ) = 0. However, these random variables are clearly not independent. The conditional distribution Y |X is not independent of X - in fact, Y |X is degenerate around X = x^2 , so the value of X is quite informative of the value of Y !!!
(3) She should not necessarily be concerned. Let X = 1 denote the event that the person has the less serious disease, and Y = 1 denote the same event for the more serious disease. Suppose that the joint probability distribution for X and Y is given by
Y = 0 Y = 1
X = 0 .075.
X = 1 .9.
From this table, it is clear that Pr(X = 1|Y = 1) =. 02 /.025 =. 8 , so this is consistent with the information given. However, Pr(Y = 1|X = 1) =. 02 /. 92 ≈. 022. So, the person is still quite unlikely to have the serious disease Y. The answer depends on the marginal probabilities associated with both Y and X.
(4)Let Cˆ denote the box that is chosen by the contestant and let C denote the actual location of the prize. Let M denote the box that the game show host (whom we will call “Monte”, given that this is the actual situation faced by real game show contestants in the TV show Let’s make a deal, hosted by Monte Hall) actually reveals as being empty. Without loss of generality, let us suppose that Cˆ = 1 and M = 2. We also assume
Pr( Cˆ = j) = 1/ 3 , j = 1, 2 , 3.
and C and Cˆ are independent.
These assumptions are reasonable - the contestant randomly chooses among the three doors, and the choice made is independent of the actual location of the prize. The second of these assumptions implies Pr( Cˆ = j|C) = Pr( Cˆ = j) = 1/ 3 , j = 1, 2 , 3.
To evaluate if the contestant should switch her choice, we need to consider and quantify the follwoing two probabilities:
Pr(C = 3|M = 2, Cˆ = 1) and Pr(C = 1|M = 2, Cˆ = 1).
Since
Pr(C = 3|M = 2, Cˆ = 1) = [Pr(M = 2, Cˆ = 1|C = 3)Pr(C = 3)]/Pr(M = 2, Cˆ = 1).
and
Pr(C = 1|M = 2, Cˆ = 1) = [Pr(M = 2, Cˆ = 1|C = 1)Pr(C = 1)]/Pr(M = 2, Cˆ = 1).
it follows that
Pr(C = 3|M = 2, Cˆ = 1) Pr(C = 1|M = 2, Cˆ = 1)
= Pr(M^ = 2,^
Cˆ = 1|C = 3)
Pr(M = 2, Cˆ = 1|C = 1) = Pr(M^ = 2|^ Cˆ = 1, C = 3)Pr( Cˆ = 1|C = 3) Pr(M = 2| Cˆ = 1, C = 1)Pr( Cˆ = 1|C = 1) = Pr(M^ = 2|^
Cˆ = 1, C = 3)
Pr(M = 2| Cˆ = 1, C = 1)
The numerator of this last expression must be 1, since Monty has no choice but to reveal box 2 if the contestant actually selected box 1 and the location of the prize is in box 2. We are not certain
An extension of this result is known as the Hammersley-Clifford Theorem in statistics, which states that an arbitrary joint density can be represented in terms of its set of conditional densities. This requires an additional condition known as positivity, which is related to the existence of the integral above and is satisfied in most cases. (Positivity basically says that whenever the marginals place positive probability over outcomes x 0 and y 0 then the joint probability of the pair (x 0 , y 0 ), also has positive probability. This rules out cases like x = y, for example, as some of you mentioned to me in class).
Since I was not at all clear regarding what conditions were needed in the derivation, you should get full credit for any reasonable attempt at this question. (If you still have questions or concerns regarding this, please come and talk with me). The bottom line is that under typical conditions, the conditionals are sufficient to define the joint, though notable exceptions do exist.
