STAT 3023/3923
Semester 2 Statistical Inference 2021
Week 3 Tutorial Solutions
1. Let X and Y be two independent standard normal random variables.
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(a) The joint density of the random vector X = (X, Y ) is
f(x,y)= 1e−12(x2+y2), −∞
as the Jacobian determinant of the transformation is
cos(θ) −r sin(θ)
det(J ) = det sin(θ) r cos(θ) = r. (c) The marginal density of R is
and the density of θ is
gR(r) = re−r2/2, r > 0. gθ(u) = 1 , 0 < u < 2π.
(d) The random variables R and θ are independent, θ ∼ U(0,2π) and R has density
fR(r) = re−r2/2, r > 0.
(e) LetV =R2. ThenP(V ≤v)=P(R≤√v).ThusV hasdensity
fV (v) = gR(√v)12v−1/2 = 12e−v/2, v > 0, an exponential distribution.
2. Let X = (X, Y ) be a random vector with joint density f(x,y)=exp(−(x+y)), x>0, y>0.
(a) Let W1 = X + Y and W2 = X − Y . We can write X = (W1 + W2)/2 and Y = (W1 − W2)/2.
J = ∂ ( w , w ) =
so |det(J)| = 12.
Thus W has joint density
|v|2 2 3. Consider the following joint probability function
(pλ)xe−pλ ((1 − p)λ)n−xe−(1−p)λ P (X = x, N = n) = x! (n − x)! ,
where λ > 0 is a fixed parameter and x = 0,1,..,n : n = 0,1,… (a)
P(N=n) = P(X=x,N=n) x=0
= (pλ)x[(1 − p)λ]n−xe−λ
g(w1,w2) = e−w1 12, w1 > 0, |w2| < w1. (b) The marginal density of W1 = X + Y is
fW1(w)= w 1e−wdw2 =we−w, w>0. −w 2
The marginal density of W2 = X − Y is
fW2(v) = ∞ 1e−wdw = 1e−|v|,−∞ < v < ∞.
a Poisson distribution with mean λ.
(b) The conditional distribution of {X|N = n} is
P (X = x, N = n) P(X=x|N=n)= P(N=n)
so the conditional distribution is B(n, p). (c)
= x px(1−p)n−x,
= e−λλn/(n!), n = 0,1,...
E(X|N = n) = np.
4. (X,Y) is uniformly distributed on the triangle x ≥ 0, y ≥ 0, x+y ≤ 2. The joint density is
f(x,y)=21, x≥0,y≥0,x+y≤2. The marginal density of X is
fX(x)= 2−x 1dy=1−x/2, 0≤x≤2. 02
The conditional density of Y given X = x is
fY|X(y)= 1/2 = 1 , 0≤y≤2−x.
1−x/2 2−x 2−x 1
E(Y |X = x) = y
5. Let (X, Y ) have joint density
f(x, y) = 12x3e−x(1+y), x > 0, y > 0.
The best predictor of Y given X is E(Y |X). The marginal density of X is ∞ 1 2 −x
fX(x)= f(x,y)dy= xe , x>0. 02
fY|X(y)=f(x,y)=xe−xy, y>0. fX (x)
dy = 1 − x/2.
The conditional density of Y given X = x is an exponential distribution and so E(Y |X) = 1/X.
6. The random vector X = (X, Y ) has joint density
f(x,y)= 9(1+x+y) , x≥0, y≥0.
2(1 + x)4(1 + y)4 (a) The marginal density of X is
= 9 ∞ x + 1 dy 2(1+x)4 0 (1+y)4 (1+y)3
= 3(2x+3), x>0. 4(1 + x)4
(b) The conditional density of Y given X = x is
fY|X(y)= f(x,y) = 6x + 6 .
fX(x) (1+y)4(2x+3) (1+y)3(2x+3)
∞ x+3 E(Y |X = x) = yfY |X (y)dy = .
0 2x+3 X 3 3−1
X=1∼N2 , . X2 2−12
(a) LetW=X1+aX2 andconsiderCov(X2,W).
Cov(X2, W ) = Cov(X1, X2) + aVar(X2) = −1 + 2a.
If a = 1/2 then X2 and W are uncorrelated and hence independent as they are obtained from a linear combination of components of a bivariate normal random vector.
(b) Note X1 = W − 1/2X2 and W and X2 are independent, so
Var(X1|X2 = 3) = Var(W ) = Var(X1) + Var(X2)/4 + Cov(X1, X2) = 5/2.
(c) The conditional distribution of X1 given X2 = 3 is the distribution of W − 3/2, as W and X2 are independent. The conditional distribution is
N (4 − 3/2, 5/2) = N (5/2, 5/2),
as E(W) = 4. Thus
P(X1 ≤2|X2 =3)=P(Z ≤ )=Φ(−1/ 10)≃0.376.
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