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University of Illinois Fall 2003
ECE 313: Solutions to Problem Set
- (a) E[R] =
∫ (^1) 0 u^ ·^2 u du^ =^
2
(b) E[V ] =
∫ (^1) 0
4 πu^3 3 ·^2 u du^ =^
8 π
Observe that E[V ] = E[^4 πR
3 3 ] =^
8 π 15 6 =^
4 π(E[R])^3 3 =^
32 π
This illustrates the general result that E[g(X)] is not necessarily equal to g(E[X]).
- Integrating both sides of λ(t) = dtd F^ (t) 1 −F (t) , we have
F (t) = 1 − exp
{ −
∫ (^) t
0
λ(t) dt
}
. (Eqn. (5.4), Pg. 216, Ross)
Evaluating this equation at t = ∞ yields ∫ (^) ∞
0
λ(t) dt = − ln(1 − F (∞)) = − ln(1 − 1) = ∞ 6 = 1.
Since the integral of λ(t) does not satisfy a criterion of a CDF (F (∞) = 1), it is clearly not a probability density function.
FX (t) = 1 − exp
{ −
∫ (^) t
0
λ(t) dt
}
= 1 − e−at
(^3) / 3 , t ≥ 0.
Differentiating both sides, we have
fX (t) = at^2 e−at
(^3) / 3 , t ≥ 0.
- The probability that an 40-year old smoker will survive, without contracting lung cancer (wclc), until age B > 40, is
P {40-year-old smoker reaches age B wclc} = P {(smoker’s lifetime wclc > B)|(smoker’s lifetime wclc > 40)}
=
1 − Fsmoker(B) 1 − Fsmoker(40)
exp
{ −
∫ (^) B 0 λ(t)^ dt
}
exp
{ −
∫ (^40) 0 λ(t)^ dt
}
= exp
{ −
∫ (^) B
40
λ(t) dt
}
= exp
{ − 0 .027(B − 40) − 0. 00025
(t − 40)^3 3
∣∣ ∣∣
B
40
} .
Plugging in the values of B,
(a) When B = 50, P {40-year-old smoker reaches age B wclc} = exp
{ − 0. 27 − 0. 325
} ∼
(b) When B = 60, P {40-year-old smoker reaches age B wclc} = exp
{ − 0. 54 − (^23)
} ∼
(a) X is a continuous RV with CDF FX (u). Because Y = FX (x), we know that it is bounded between 0 and 1. Assuming FX (x) is strictly monotonic,
FY (v) = P {Y ≤ v} = P {FX (x) ≤ v}
(a) = P {(X) ≤ F (^) X− 1 (v)}
(b) = FX (F (^) X− 1 (v)) = v.
Therefore, Y is a uniform RV in the interval (0, 1). Equalities (a) and (b) follow because FX (u) is a strictly monotonic function and hence, one-to-one. Therefore, for all values of v ∈ (0, 1), FX (u) is well defined: also, FX (F (^) X− 1 v) = v because of the existence of the inverse. However, if Y is not strictly monotonic, but non-decreasing, then there can exist intervals I such that FX (x) = FX (y) = c, ∀ x, y ∈ I. Consequently, FX (x) is a many-to-one mapping and equality (a) is invalid as the inverse F (^) X− 1 (·) would not exist. To overcome this problem, we define a new function g(y)
4 = inf {x : FX (x) = y}. It follows that
FX (g(v)) = v (1) and g(FX (x)) ≤ x. (2)
Thus P {FX (X) ≤ v}
(a) = P {FX (X)) ≤ FX (g(v))}
(b) = P {X ≤ g(v)} = FX (g(v)) = v. Equality (a) follows from (1),while equality (b) holds since FX (·) is non-decreasing (FX (a) ≤ FX (b) ⇔ a ≤ b). (b) For F−^1 (·) to exist, F(·) has to be strictly monotonically increasing, so that we have
FX (u) = P {X ≤ u} = P {F−^1 (Y ) ≤ u}
(a) = P {(Y ) ≤F(u)}
(b) =F(u).
Equality (a) follows because F is monotone increasing. Equality (b) follows because Y is a uniform RV in [0, 1]: care must be taken, though, to ensure that F(u) ∈ [0, 1], but since F(u) is a CDF, this holds trivially for all u. Therefore, we have generated an RV X with the desired CDF F(u) for all u. For the more general case of a non-decreasing F(·), we can construct a “pseudo- inverse” g(·), as in part (a), to approximate F−^1 (·). The proof that X has CDF F(x), even in this case, follows along the same lines as part (a). Alternative Proof :
(a) Assuming FX (x) is a strictly increasing, differentiable function of x, we can apply Theorem 7.1, Pg.224, Ross, to evaluate the pdf of Y:
fY (y) = fX [F (^) X− 1 (y)] ·
∣∣ ∣∣^ d dy
F (^) X− 1 (y)
∣∣ ∣∣, for all y ∈ [0, 1]
where d dy
F (^) X− 1 (y) =
dx dy
dy dx
d dx FX^ (x)
fX (x)
Therefore,
fY (y) = fX [F (^) X− 1 (y)]
∣∣ ∣∣^1 fX (x)
∣∣ ∣∣ = fX (x)
∣∣ ∣∣^1 fX (x)
∣∣ ∣∣ = 1 since fX (x) ≥ 0 , ∀ x
or, Y ∼ U [0, 1].
E[Z] is minimized with respect to α when dE dα[Z ]= 2α − 2
√ 2 /π = 0, or αmin =
√ 2 /π.
fX (x) =
1 2 π ,^ −π^ ≤^ x^ ≤^ π,
0 , otherwise.
For 0 ≤ y ≤ 1, the event {Y ≤ y} satisfies
{Y ≤ y} = {sin X ≤ y} = {π − sin−^1 y < X ≤ π}
⋃ {−π < X ≤ sin−^1 y}.
Since these two events are disjoint, we obtain FY (y) = FX (π) − FX (π − sin−^1 y) + FX (sin−^1 y) − FX (−π). Hence
fY (y) =
dFY (y) dy
= fX (π − sin−^1 y)
1 − y^2
1 − y^2
=
π
1 − y^2
0 ≤ y < 1.
The pdf derived above is valid over the half-open interval [0,1) only, and not [0,1]. This is because the CDF is not differentiable at 1, −1. However, it is continuous, which implies the pdf is finite at these points and can be assigned any finite value, in accordance with the properties of a continuous CDF. If this calculation is repeated for − 1 ≤ y ≤ 0, and for |y| > 1, we obtain the complete solution:
fY (y) =
1 π
√^1
1 −y^2
, |y| < 1 ,
0 , otherwise.
Alternative Method: y = g(x) = sin x is a differentiable function of the real variable x.
sin−1Y π−sin−1Y
Y −π π
0 X
sin X
1/π
fY(Y)
−1 (^01) Y (a) (b)
Figure 1: (a) Roots of y = sin x; (b) the corresponding probability density function fY (y).
The roots of y − sin x for y > 0 are x 1 = sin−^1 y and x 2 = π − sin−^1 y. Also,
dg(x) dx
= cos x,
which must be evaluated at the two roots x 1 and x 2. At x 1 = sin−^1 y we get dg/dx|x=x 1 = cos(sin−^1 y). Likewise, when x 2 = π − sin−^1 y, we get
dg dx
∣∣ ∣∣ x=x 2
= cos(π − sin−^1 y) = cos π cos(sin−^1 y) + sin π sin(sin−^1 y)
= − cos(sin−^1 y).
Let θ = sin−^1 y. Then y = sin θ and cos(sin−^1 y) = cos θ =
1 − y^2. Hence ∣∣ ∣∣^ dg dx
∣∣ ∣∣ x=x 1
∣∣ ∣∣^ dg dx
∣∣ ∣∣ x=x 2
√ 1 − y^2.
Finally,
fX (sin−^1 y) = fX (π − sin−^1 y) =
1 2 π ,^0 ≤^ y^ ≤^1
0 , y > 1
Using these results in the general formula
fY (y) =
∑^ n
i=
fX (xi)/|g′(xi)|, xi = xi(y), g′(xi) 6 = 0,
we have
fY (y) =
1 π
√^1
1 −y^2 0 ≤^ y <^1 ,
0 , y > 1.
Note that at y = 1, g′(x)|(y=1) = g′(π/2) = 0, so the formula is not applicable here. However, P (Y = 1) = P (X = π/2) = 0 (X is a continuous RV). Therefore, FY (y) is continuous at y = 1, so fY (y) can be assigned any finite value. Proceeding similarly for y < 0, we arrive at the same result as Eqn. (4) above.
fX (x) =
1 2 ,^0 ≤^ x^ ≤^2 ,
0 , otherwise.
From the definition of g(·), we see that Y can only assume values in the interval [0, 1]. Therefore, FY (y) = 0, y < 0 , and FY (y) = 1, y > 1. For 0 ≤ y ≤ 1 the event {Y ≤ y} satisfies
{Y ≤ y} = { 2 X ≤ y, 0 ≤ X ≤ 1 / 2 }
⋃ {(2 − 2 X) < y, 1 / 2 ≤ X ≤ 1 }
⋃ { 0 ≤ y, X ≥ 1 }
= { 0 < X < y/ 2 }
⋃ {(1 − y/2) ≤ X ≤ 1 }
⋃ { 0 ≤ y, X ≥ 1 }.
Since the events on the previous line are disjoint, we obtain
FY (y) = FX (y/2) − FX (0) + FX (1) − FX (1 − y/2) + FX (2) − FX (1) = 1 / 2 · (y/ 2 − 0 + 1 − 1 + y/2 + 1) = (y + 1)/ 2 , 0 ≤ y ≤ 1.
