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Fall 2018S&DS 241: Probability Theory with Applications Homework 9Due: Nov 18, 2018 in class Prof. Yihong WuSolutions by Ganlin Song. LetT∈[0, 1 /2] be the distance between the midpoint of the needle and the closestline and let θ∈[0, π/2] be the acute angle between the needle and the lines. ...

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Fall 2018 S&DS 241: Probability Theory with Applications Homework 9 Due: Nov 18, 2018 in class Prof. Yihong Wu Solutions by Ganlin Song. 1. Let T ∈ [0, 1/2] be the distance between the midpoint of the needle and the closest line and let θ ∈ [0, π/2] be the acute angle between the needle and the lines. The needle touches at least one line if and only if T ≤ sin(θ). Moreover, the needle touches exactly one line if and only if T ≤ sin(θ) and T ≤ (1 − sin(θ)). Let L be the number of lines the needle intersects. Then L ∈ {0, 1, 2, 3}. (a) The probability that the needle touches at least one line is equal to P (L ≥ 1) = P (T ≤ sin(θ)). Since (T, θ)′ has the joint density ( 4/π for 0 ≤ t ≤ 1/2, 0 ≤ θ ≤ π/2 fT ,θ (t, θ) = , 0 otherwise it follows that P (L ≥ 1) = P (T ≤ sin(θ)) Z π/2 Z min{1/2, sin(θ)} = (4/π)dtdθ =

Z

=

0 π/6

0

√ 4−2 3 π

0

Z

sin(θ )

(4/π)dtdθ +

Z

π/2 Z 1/2

π/6

0

(4/π)dtdθ

0

+ 32 .

(b),(c) To find the expected number of intersections, we first find the probability that the needle touches exactly one line. This is equal to P (L = 1) = P (T ≤ sin(θ), T ≤ 1 − sin(θ)) Z π/2 Z min{1/2, sin(θ), 1−sin(θ)} = (4/π)dtdθ 0

=

Z

0

= =

0 π/6 Z sin(θ ) 0

√ 4−2 3 + 34 π √ 4 + 4−4π 3 3

−

(4/π)dtdθ +

Z

π/2 Z 1−sin(θ )

π/6

√ 2 3 π

(4/π)dtdθ

0

√

Note that P (L = 0) = 1 − P (L ≥ 1) = 1 − ( 4−2π 3 + 23 ), P (L = 2) = P (L ≥ 1) − P (L = 1) = √ √ √ ( 4−2π 3 + 32 ) − ( 43 + 4−π4 3 ) = 2 π3 − 32 , and P (L = 3) = 0. Thus, E(L) = 0 · P (L = 0) + 1 · P (L = 1) + 2 · P (L = 2) + 3 · P (L = 3) = P (L = 1) + 2P (L = 2)

= ( 34 + = π4 .

√ 4−4 3 ) π

√

+ 2( 2 π 3 − 32 )

1

Alternatively, we can cut the needle in half to produce two pieces of length 1. We already know that if the needle has length 1, then the expected number of intersections is 2π . Thus, by linearity of expectation E(L) = π2 + 2π = π4 . Thus, P (L = 1) + 2P (L = 2) = 4π . This enables us to find √ the distribution of L since by part (a), we know that P (L = 1) + P (L = 2) = 4−2π 3 + 32 . 2. (a) Note that c must satisfy

Now, (b)

R1 R1

Z

1

−1

−1 −1 c(1

Z

1

c(1 + xy)dxdy = 1.

−1

+ xy)dxdy = 4c. Thus 4c = 1 or c = 1/4. E(Z) =

Z

E(Z 2 ) =

Z

1

1

Z

xyc(1 + xy )dxdy =

−1 −1 1

−1

Z

1

Z

x2 y2 c(1 + xy)dxdy =

−1

1 −1

Z

Z

1

1

Z

cx2 y2 dxdy =

−1

−1

1

cx2 y2 dxdy =

−1

V ar(Z) = E (Z 2 ) − (E (Z))2 =

1 9 1 9

8 81

(c) We will first calculate the marginal densities fX and fY and show that fX,Y 6= fX fY . Now, Z 1 fX (x) = fX,Y (x, y)dy −1 1

=

Z

(1/4)(1 + xy)dy

−1

= 1/2, for |x| < 1 and fX (x) = 0 otherwise. By symmetry fX = fY . It is clear that fX,Y 6= fX fY . (d) We will calculate FX 2 ,Y 2 (s, t): FX 2 ,Y 2 (s, t) = P (X 2 ≤ s, Y 2 ≤ t) √ √ = P (|X| ≤ s, |Y | ≤ t) √ √ √ √ = P (X ≤ s, |Y | ≤ t) − P (X < − s, |Y | ≤ t) √ √ √ √ = P (X ≤ s, Y ≤ t) − P (X ≤ s, Y < − t)− √ √ √ √ [P (X < − s, Y ≤ t) − P (X < − s, Y < − t)] √ √ √ √ = P (X ≤ s, Y ≤ t) − P (X ≤ s, Y ≤ − t)− √ √ √ √ [P (X ≤ − s, Y ≤ t) − P (X ≤ − s, Y ≤ − t)] √ √ √ √ = FX,Y ( s, t) − FX,Y ( s, − t)− √ √ √ √ FX,Y (− s, t) + FX,Y (− s, − t) 2

2

∂ ∂ F Next, we use the fact that fX 2 ,Y 2 (s, t) = ∂s∂t FX 2 ,Y 2 (s, t) and fX,Y (s, t) = ∂s∂t X,Y (s, t) to conclude that √ √ fX 2 ,Y 2 (s, t) = 4√1st [fX,Y ( s, t)+ √ √ √ √ √ √ fX,Y ( s, − t) + fX,Y (− s, t) + fX,Y (− s, − t)]

=

√1 , 4 st

2

for 0 < s < 1 and 0 < t < 1 and fX 2 ,Y 2 (s, t) = 0 otherwise. Since fX 2 (s) = fY 2 (s) = 2

1√ , 2 s

2

follows that fX 2 ,Y 2 = fX 2 fY 2 and hence X and Y

are independent. √ √ Alternatively, we can evaluate FX 2 ,Y 2 (s, t) = P (|X| ≤ s, |Y | ≤ t) as the integral Z

√ t

√ − t

Z

√ s

f (x, y)dxdy √ X,Y − s

√ t

Z

=

√ − t

Z

√ s

(1/4)(1 + xy)dxdy =

√ − s

√ √ s t,

which implies independence. 3. (a) Using the hint, note that FZ (z) = P (Z ≤ z)

= P (X/Y ≤ z, Y > 0) + P (X/Y ≤ z, Y ≤ 0) = P (X ≤ zY, Y > 0) + P (X ≥ zY, Y ≤ 0) Z 0 Z ∞ Z ∞ Z zy φ(x)φ(y)dxdy + φ(x)φ(y)dxdy = = =

Z

0 ∞

0 Z ∞

−∞ 0

−∞ zy

Z (

Z

φ(x)dx)φ(y)dy +

−∞

Φ(zy)φ(y)dy +

0

0

Z

−∞

−∞

zy

Z (

∞

φ(x)dx)φ(y)dy

zy

(1 − Φ(zy))φ(y)dy,

as desired. (b) Next, fZ (z) = = =

d dz FZ (z) ∞ d dz Φ(zy)φ(y)dy 0 Z ∞

Z

0

+

yφ(zy)φ(y)dy −

Z

Z

0

−∞ 0

d dz (1

− Φ(zy))φ(y)dy

yφ(zy)φ(y)dy. −∞

By a change of variables u = −y and the fact that φ is an even function, we have that Z 0 Z ∞ yφ(zy)φ(y)dy = yφ(zy)φ(y)dy. 0

−∞

Thus, fZ (z) = 2 = =

Z

1 π

∞

Z0

yφ(zy)φ(y)dy

∞

ye−(z

2 +1)y 2 /2

dy

0 1 , π(z 2 +1)

for all z ∈ R. (c) The value of E|Z| is positive infinity. This can be seen from the estimate Z ∞ Z Z ∞ Z ∞ Z ∞ |z|dz 2zdz 2zdz 2zdz 1 E|Z| = ≥ ≥ = = π π(z 2 +1) π(z 2 +1) π(z 2 +1) π(z 2 +z 2 ) −∞

0

1

1

3

1

∞

dz z

= +∞.

it

4. See below.

5. (a) Let θ1 be the angel you choose going away from the bank. Then θ1 ∼ U (0, π). The distance from the bank D = sin θ1 . Then the expectation of D is Z π 1 2 (sin θ1 ) dθ1 = . E[D] = π π 0 4

(b) Let θ2 be the angel you choose to return to the bank. Then θ2 ∼ U (0, 2π). Condition on θ1 , we have π  o n    P {return to the bank|θ1 } = P θ2 ∈ 0, 2  − θ1  . 2 Therefore, P {return to the bank} =

Z

0

π

1 P {return to the bank|θ1 } dπ = π

5

Z

π 0

Z

2|π/2−θ1 | 0

1 1 1 dθ2 dθ1 = . 4 2π π...