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Math 431 – Spring 2008
Solutions 12
- Problems from Chapter 6:
- Although one could use the formula for the joint density of functions of random variables, it is actually easier to compute this directly. We calculate the joint dis- tribution function first:
FR,Θ(r, θ) = P(R ≤ r, Θ ≤ θ) = P(
X^2 + Y 2 ≤ r, tan−^1 (Y /X) ≤ θ)
{R ≤ r, Θ ≤ θ} describes the points in polar coordinate for which the radius is at most r and the angle is in [−π/ 2 , θ] ∪ [π/ 2 , π + θ]. Since we have a uniform distribution on the unit disk, the probability (which is the are divided by π) is
FR,Θ(r, θ) = r^2 (π/2 + θ)
π Differentiating this wrt r and θ gives the joint density:
f (r, θ) =
2 r π
Note: if instead of tan−^1 Y /X we use the angle of (X, Y ) (i.e. a number in [0, 2 π]) then we get r/π instead.
- a) We use the formula covered in class.
g 1 (x, y) = xy, g 2 (x, y) = x/y h 1 (u, v) =
uv, h 2 (u, v) =
u/v
(we can take the positive square roots as X, Y ≥ 1. Also
∂xg 1 = y, ∂yg 1 = x, ∂xg 2 = 1/y, ∂yg 2 = −x/y^2
and J(x, y) =
y x 1 /y −x/y^2
∣ =^ −^2 x/y
Thus fU,V (u, v) = f (x, y)|J(x, y)|−^1 =
x^2 y^2
y 2 x
2 u^2 v Here u, v must satisfy uv ≥ 1 , u/v ≥ 1 and u, v > 0.
E(|x| + |y|) = E|x| + E|y| = 2E|x| = 2
− 1. 5
|x|dx = 3/ 2
- The expectation of the sum is the sum of the expectations 10 × 7 /2 = 35.
- b) Let Ij be the indicator of the event that the jth^ object is not chosen by either A or B. Then we need the expectation of X = I 1 + I 2 +... I 10. But
EIj = P(j is not chosen by either A or B) =
[ ( 9
3
3
] 2
so EX = 10 × .49 = 4. 9
- a) X = I 1 + I 2 +... In where Ij is the indicator that the jth^ urn is empty.
EIj = P(all the balls go to the other urns) = (n − 1)n/nn
which give EX = nEI 1 = (n − 1)n/nn−^1.
- b) X = I 1 + · · · + In where Ij is the indicator that the jth^ man has a women next to him.
EIj = P( the jth^ man has a women next to him) = 1 − P( the jth^ man doesn’t have a women next to him)
= 1 −
(n − 1)(n − 2)((2n − 3)!) (2n − 1)!
n − 2 2(2n − 1)
3 n 4 n − 2
which gives EX = 3 n
2 4 n− 2. Note: to compute the probability we note that the size of the sample space is (2n − 1)! (it is enough to order the 2n − 1 other persons) and the size of the event is (n − 1)(n − 2)((2n − 3)!) (choose a man to the left, a man to the right neighbor and order the other 2n − 3 persons).
- Theoretical Exercise 1 from Chapter 7:
E(x−a)^2 = E(X−EX+EX−a)^2 = E(X−EX)^2 +E(EX−a)^2 +E [(X − EX)(EX − a)]
The first term is V ar[X], the second term is a just the number (EX − a)^2 and the third term is 0 because
E [(X − EX)(EX − a)] = (EX − a)E[X − EX] = (EX − a)0 = 0.
Thus E(x − a)^2 = V ar[X] + (EX − a)^2 ≥ V arX = E(X − EX)^2
which shows that the minimum is achieved at a = EX. Alternate solution: E(X − a)^2 = a^2 − (2EX)a + EX^2
This is a quadratic function in a which we can minimize by finding the critical values. The derivative is 2a − 2 EX so only critical value is a = EX which is a global minimum (since the second derivative is 2 which is positive.
