Download 1. (a) (b) , Since for all x, y, x and Y are independent variable 2. W : a

1.
−2(x+y)
fXY (x, y) = {Ae
∞
∞
∞
(a) 1 = ∫0 ∫0 Ae−2(x+y) dxdy = ∫0 Ae−2y
∞
if x ≥ 0, y ≥ 0
otherwise
∞
A ∞
A
(∫0 e−2x dx)dy = ∫0 e−2x dx = , A = 4
0
2
∞
4
(b) fy (y) = ∫0 4e−2(x+y) dx = 4e−2y ∫0 e−2x dx = 2e−2y , fx (x) = 2e−2x
Since f(x, y) = fX (x)fy (y) for all x, y, x and Y are independent variable
2.
W : a soccer team will win a game
S : the soccer team will not have a player sent off
P(W) = 0.55 P(S) = 0.9 P(W|S) = 0.6
The probability that team has a player sent off but still wins the game :P(W ∩ S c )?
P(W) = P(W|S c )P(S c ) + P(W|S)P(S) = 0.54 + P(W ∩ S c ) = 0.55 P(W ∩ S c ) = 0.01
3.
(a) xi is a variable with mean μ and variance σ2 , by Chebyshev’s inequality, P{|xi − μ| ≥ kσ} ≤
Given E(x̅) = μ , Var(x̅) =
cσ
√n
} ≤ lim
1
n→∞ √n
( ϵ)
σ2
n
, we obtain that P {|x̅ − μ| ≥
cσ
√n
}≤
1
where c =
c2
√n
ϵ . lim P {|x̅
σ
n→∞
1
k2
.
− μ| ≥
→0
2
σ
(b) Let x1 ⋯ xn be a sequence of independent and identically distributed random variables each
having mean μ and variance σ2 . By central limit theorem, for n large, the distribution of x1 + ⋯ +
xn is approximately normal with mean nμ and variance nσ2 . For n large, it follows that
x1 +⋯+xn −nμ
σ√n
=
x̅−μ
σ√n
= Y is approximately a standard normal random variable (5 point)
t
ϕY (t) = E(etY ) = E (e√n
(1 +
t
2√n
n
t2
2
) =e
∑ Yi
t
(Y1 + ⋯Yn )
) = E (e√n
t
(Y1 )
) = E (e√n
n
) = (1 +
t
√n
E(Y1 ) +
t2
2√n
n
E(Y1 )2 ) ≈
as n → ∞ . So we obtain uniquely moment generating function. (5 point)
(c) A random variable is
(n−1)s2
σ2
μ)2 + (x̅ − μ)2 − 2(xi − μ)(μ − x̅)) =
=
(n−1)s2 1
σ2
n−1
1
n
∑
((x
i−
σ2 i=1
∑ni=1(xi − x̅)2 =
1
σ2
μ)2 − (x̅ − μ)2 ) =
1
∑n ((xi
σ2 i=1
(x̅−μ)2
~χ2n − χ12 ~χ2n−1
σ2
∑ni=1(xi − μ + μ − x̅)2 =
2
∑n
i=1(xi −μ)
−
σ2
−
,
since xi is a normal distribution having mean μ and variance σ2 . (8 point)
You should include proscess ∅χ2n (t) = ∅χ2n−1+χ21 (t) = ∅χ2n−1 (t)∅χ21 (t). (2 point)
(d) A random variable is T =
2
x̅−μ
(s )2
⁄
√n
=
̅−μ)2
n(x
σ2
s2
σ2
=
̅−μ)2
(x
(σ⁄ )2
√n
(n−1)s2
σ2 (n−1)
= χ2
z2
n−1⁄
n−1
= χ2
n−1⁄
approximately a standard normal random variable.
4.
f (xi )
, we obtain
(xi )|
(−y−μ)2
(y−μ)2
By using fY (y) = ∑ni=1 |gx′
Thus fY (y) =
1
√2πσ
exp (−
2σ2
−
2σ2
fY (y) = fX (−y) − fX (y) for y ≥ 0.
) for y ≥ 0
χ2
1⁄
1
n−1
= F1,n−1 , since
x̅−μ
σ√n
is
5.
To obtain third head on the 9th roll, exactly 8th times with second head is obtained
8 1 1
7
P(x = 9) = ( ) ( )6 ( )3 =
2 2 2
128
6.
X~B(400, 0.9)
If n is sufficiently large, X can approximate N(np, np(1 − p)) i. e X~N(360, 36)
Then P(X ≤ 372) = P (
X−360
6
≤ 2) = P(Z ≤ 2) = ϕ(2)(point 9). If you include +0.5 or –0.5(point 1)
7.
(a)
N(t) be a Poisson process with λ = 12. N(t) that has a Poisson distribution with mean λt is the
number of events occurring within such a time interval. The probability that there is no visit within
1
20
6
0!
a 10 minutes interval is P (N ( ) = 0) = e−2
= e−2
(b)
X1 ⋯ Xn are i.i.d exponential distribution with a parameter λ = 12 i. e. ϕXi = (
⋯ + Xk is a gamma distribution with parameters α = k and λ. i. e. ϕYk = (
λ
λ−t
λ
λ−t
k
). And Yk = X1 +
) . So Y10 ~ G(10,12).
1
1
To express with χ2α,n , we can get ̂
Y~ G(10, ) by changing λ = 12 per an hour to λ̂ = per 2.5
2
2
̂ ≥ 24) = P(χ2α,20 ≥ 24)
minutes. Then α = P(Y10 ≥ 1) = P(Y
8.
The likelihood function is given by
n
e−λ λx1 e−λ λxn e−nλ λ∑1 xi
⋯
=
x1 !
xn !
x1 ! ⋯ xn !
Thus, logf(x1 ⋯ xn |λ) = −nλ + ∑n1 xi logλ − logc where c = ∏ni=1 xi ! does not depend on λ, and
∑n1 xi
d
logf(x1 ⋯ xn |λ) = −n +
dλ
λ
∑n x
By equating ti zero, we obtain that the maximum likelihood estimate λ̂ equals λ̂ = 1 i
f(x1 ⋯ xn |λ) =
n