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ECE361 Fall 2002 Solutions to Homework 1
- 4.6 of text book PS: 1) X can take four different values. 0, if no head shows up, 1, if only one head shows up in the four flips of the coin, 2, for two heads and 3 if the outcome of each flip is head.
- X follows the binomial distribution with n = 3. Thus
P(X = k) =
( 3 k
) pk(1 − p)^3 −k^ for 0 ≤ k ≤ 3 0 otherwise
FX (k) =
∑^ k m=
( 3 m
) pm(1 − p)^3 −m
Hence
FX (k) =
0 k < 0 (1 − p)^3 k = 0 (1 − p)^3 + 3p(1 − p)^2 k = 1 (1 − p)^3 + 3p(1 − p)^2 + 3p^2 (1 − p) k = 2 (1 − p)^3 + 3p(1 − p)^2 + 3p^2 (1 − p) + p^3 = 1 k = 3 1 k > 3
CDF
(1−p)^3 -1 0 1 2 3 4
P(X > 1) =
∑^3 k=
( 3 k
) pk(1 − p)^3 −k^ = 3p^2 (1 − p) + (1 − p)^3
- 4.8 of text book PS: 1) Since limx→∞ FX (x) = 1 and FX (x) = 1 for all x ≥ 1 we obtain K = 1.
- The random variable is of the mixed-type since there is a discontinuity at x = 1. lim→ 0 FX (1 − ) = 1/2 whereas lim→ 0 FX (1 + ) = 1
P(
< X ≤ 1) = FX (1) − FX (
P(
< X < 1) = FX (1−) − FX (
P(X > 2) = 1 − P(X ≤ 2) = 1 − FX (2) = 1 − 1 = 0
- 4.10 of text book PS: 1) The random variable X is Gaussian with zero mean and variance σ^2 = 10−^8. Thus P(X > x) = Q( xσ ) and
P(X > 10 −^4 ) = Q
( 10 −^4 10 −^4
) = Q(1) =. 159
P(X > 4 × 10 −^4 ) = Q
( 4 × 10 −^4 10 −^4
) = Q(4) = 3. 17 × 10 −^5
P(− 2 × 10 −^4 < X ≤ 10 −^4 ) = 1 − Q(1) − Q(2) =. 8182
P(X > 10 −^4 |X > 0) =
P(X > 10 −^4 , X > 0)
P(X > 0)
P(X > 10 −^4 )
P(X > 0)
- y = g(x) = xu(x). Clearly fY (y) = 0 and FY (y) = 0 for y < 0. If y > 0, then the equation y = xu(x) has a unique solution x 1 = y. Hence, FY (y) = FX (y) and fY (y) = fX (y) for y > 0. FY (y) is discontinuous at y = 0 and the jump of the discontinuity equals FX (0).
FY (0+) − FY (0−) = FX (0) =
In summary the PDF fY (y) equals
fY (y) = fX (y)u(y) +
δ(y)
The general expression for finding fY (y) can not be used because g(x) is constant for some interval so that there is an uncountable number of solutions for x in this interval.
E[Y ] =
∫ (^) ∞ −∞ yfY (y)dy
=
∫ (^) ∞ −∞ y
[ fX (y)u(y) +
δ(y)
] dy
=
√^1
2 πσ^2
∫ (^) ∞ 0
ye−^
y^2 2 σ^2 dy = √σ 2 π
∫ (^1) 0
x^2
∣∣ ∣∣
1 y
dy +
∫ (^1) 0 yx
∣∣ ∣∣
1 y
dy
=
∫ (^1) 0
(1 − y^2 )dy +
∫ (^1) 0 y(1 − y)dy
=
P(X > Y |X + 2Y > 1) = P(X > Y, X + 2Y > 1)/P(X + 2Y > 1)
The region over which we integrate in order to find P(X > Y, X + 2Y > 1) is marked with an A in the following figure.
HH
...H
HH
HH HH x
y
1 / 3
(1,1)
x+2y=
A
Thus
P(X > Y, X + 2Y > 1) =
∫ (^1) (^13)
∫ (^) x 1 − 2 x^ (x^ +^ y)dxdy
=
∫ (^1) (^13)
[ x(x − 1 − x 2
(x^2 − ( 1 − x 2
)^2 )
] dx
=
∫ (^1) (^13)
x^2 −
x −
) dx
=
P(X + 2Y > 1) =
∫ (^1) 0
∫ (^1) 1 − 2 x^ (x^ +^ y)dxdy
=
∫ (^1) 0
[ x(1 −
1 − x 2
1 − x 2
)^2 )
] dx
=
∫ (^1) 0
x^2 +
x +
) dx
=
×
x^3
∣∣ ∣∣ 1 0
×
x^2
∣∣ ∣∣ 1 0
x
∣∣ ∣∣ 1 0 =
Hence, P(X > Y |X + 2Y > 1) = (49/108)/(7/8) = 14/ 27
- When X = Y the volume under integration has measure zero and thus
P(X = Y ) = 0
- Conditioned on the fact that X = Y , the new p.d.f of X is
fX|X=Y (x) = fX,Y (x, x) ∫ (^1) 0 fX,Y^ (x, x)dx^
= 2x.
In words, we re-normalize fX,Y (x, y) so that it integrates to 1 on the region char- acterized by X = Y. The result depends only on x. Then P(X > 12 |X = Y ) = ∫ (^1) 1 / 2 fX|X=Y^ (x)dx^ = 3/4.
fX (x) =
∫ (^1) 0 (x + y)dy = x +
∫ (^1) 0 ydy = x +
fY (y) =
∫ (^1) 0
(x + y)dx = y +
∫ (^1) 0
xdx = y +
- FX (x|X + 2Y > 1) = P(X ≤ x, X + 2Y > 1)/P(X + 2Y > 1)
P(X ≤ x, X + 2Y > 1) =
∫ (^) x 0
∫ (^1) 1 − 2 v^ (v^ +^ y)dvdy
=
∫ (^) x 0
[ 3
v^2 +
v +
] dv
=
x^3 +
x^2 +
x
Hence, fX (x|X + 2Y > 1) =
3 8 x
8 x^ +^
3 8 P(X + 2Y > 1)
x^2 +
x +
E[X|X + 2Y > 1] =
∫ (^1) 0 xfX (x|X + 2Y > 1)dx
=
∫ (^1) 0
x^3 +
x^2 +
x
)
×
x^4
∣∣ ∣∣
1 0
×
x^3
∣∣ ∣∣
1 0
×
x^2
∣∣ ∣∣
1 0
- (a) True. The intuitive way of thinking is through the following two statements:
- A and Ac^ (and similarly the pair B and Bc) have the same amount of infor- mation.
- If A and B are independent then knowing the occurrence of one event does not tell anything extra about the other.
- Putting together the above two facts, we conclude that Ac^ and Bc^ are also independent.
