Homework 6 Solutions
Stat 516
(1.7.9) Part (a): X ∼ Bin(4, 1/4), and the probabilities for the distribution can be seen in
the table below. The value of x which satisfies P (X < x) ≤ 12 and P (X ≤ x) ≥ 12 is x = 1,
so the median m = 1 .
x
p(x)
0
0.3164
1
0.4219
2
3
4
0.2109 0.0469 0.0039
Part (b): We get m = (0.5)1/3 = 0.794
Z m
m
3x2 dx = 0.5 ⇔ x3 0 ⇔ m3 = 0.5
0
Part (c): Let x = tan θ, so
−π/2 < θ < π/2.
dx
dθ
= sec2 θ. Also note that tan2 θ + 1 = sec2 θ and now
1
π
Z
m
−∞
m0
1
1
dx =
2
1+x
2
Z
1
1
1
sec2 θdθ =
2
π −π/2 sec θ
2
θ m 0
1
⇔ −π/2 =
π
2
0
⇔m =0
⇔
Since tan(0) = 0, then m = 0
(1.7.13) We have that F (x) = 1 − e−x − xe−x , 0 ≤ x < ∞. The pdf is
f (x) = F 0 (x) = e−x − e−x + xe−x = xe−x , 0 ≤ x < ∞.
To get the mode we set
f 0 (x) = 0 ⇔ e−x − xe−x = 0,
and get the mode to be x = 1 . To get the median, we set
F (x) =
1
1
⇔ e−x + xe−x = .
2
2
1
and get the median to be x = 1.678 . We can use Newton’s method with update step,
xn+1 = xn −
e−x + xe−x −
−xe−x
1
2
(1.7.20) Use theorem 1.7.1. Since 0 < x < 3, then 0 < x3 < 27. In addition, we have
1
dx
= y −2/3
x = y 1/3 ,
dy
3
dx 1
g(y) = f (y 1/3 ) = , 0 < y < 27.
dy
27
(1.7.22) We know that if − π2 < x < π2 , then −∞ < tan(x) < ∞. In addition, we have
x = tan−1 (y),
g(y) = f
1
1 + y2
dx
1
=
dy
1 + y2
dx 1
=
dy π(1 + y 2 ) , −∞ < y < ∞.
(1.10.6) We note that X has support on (0, ∞). We note Jensen’s inequality for φ convex
that
φ[E(X)] ≤ E[φ(X)]
Part (a): φ(x) = x1 , φ00 (x) =
Jensen’s inequality,
2
, 2
x3 x3
> 0 for 0 < x < ∞, therefore, φ(x) is convex on and by
E(1/X) ≥
1
E(X)
Part (b): φ(x) = − log x, φ00 (x) = x12 , x12 > 0 for 0 < x < ∞, so φ(x) convex and Jensen’s
inequality gives,
E[− log X] ≥ − log[E(X)]
Part (c): Since φ(x) = log(1/x) = − log x, then following our conclusion from part (b),
E[log(1/X)] ≥ log[1/E(X)]
Part (b): φ(x) = x3 , φ00 (x) = 6x, 6x > 0 for 0 < x < ∞, thus,
E[X 3 ] ≥ [E(X)]3
2