Jaime Frade STA5166

2. Show that

E[Y] = ∫_{0}^{∞} P {Y > y}dy − ∫_{0}^{∞} P {Y < −y}dy  (1)

E[Y] = ∫_{−∞}^{∞} xf_{y}(x)dx
= ∫_{−∞}^{0} xf_{y}(x)dx + ∫_{0}^{∞} xf_{Y} (x)dx
= − [− ∫_{−∞}^{0} xf_{Y} (x)dx] + ∫_{0}^{∞} xf_{Y} (x)dx  (2)

Show the following for (2) to hold:

∫_{0}^{∞} P {Y > y}dy = ∫_{0}^{∞} xf_{Y} (x)dx  (3)
∫_{0}^{∞} P {Y < −y}dy = − ∫_{−∞}^{0} xf_{Y} (x)dx  (4)

By definition,

P {Y > y} = ∫_{y}^{∞} f_{Y} (x)dx

Intergrating both sides,

∫_{0}^{∞} P {Y > y}dy = ∫_{0}^{∞} ∫_{y}^{∞} f_{Y} (x)dx dy  (5)

Changing the order of intergration in (5),

∫_{0}^{∞} P {Y > y}dy = ∫_{0}^{∞} ∫_{0}^{x} f_{Y} (x)dy/, dx
= ∫_{0}^{∞} (∫_{0}^{x} dy) f_{Y} (x)dx
= ∫_{0}^{∞} xf_{Y} (x)dx  (6)

By definition,

P {Y < −y} = ∫_{−∞}^{−y} f_{Y} (x)dx

Intergrating both sides,

∫_{0}^{∞} P {Y < −y}dy = ∫_{0}^{∞} ∫_{−∞}^{−y} f_{Y} (x)dx dy  (7)

Let x = −x′. Rewrite (7).

∫_{0}^{∞} P {Y < −y}dy = ∫_{0}^{∞} ∫_{y}^{∞} f_{Y} (−x′)dx′ dy  (8)

Changing the order of intergration in (8),

∫_{0}^{∞} P {Y < −y}dy = ∫_{0}^{∞} ∫_{0}^{x′} f_{Y} (−x′)dy dx′

= ∫_{0}^{∞} (∫_{0}^{x′} dy) f_{Y} (−x′)dx′

= ∫_{0}^{∞} x′f_{Y} (−x′)dx′  (9)

Let −x = x′. Rewrite (9).

∫_{0}^{∞} P {Y < −y}dy = ∫_{0}^{∞} x′f_{Y} (−x′)dx′
= ∫_{0}^{−∞} −xf_{Y} (x)(−dx)
= − ∫_{−∞}^{0} xf_{Y} (x)dx  (10)

Therefore by (6) and (10), will obtain (2).

E[Y] = − [− ∫_{−∞}^{0} xf_{Y} (x)dx] + ∫_{0}^{∞} xf_{Y} (x)dx
= ∫_{0}^{∞} xf_{Y} (x)dx − [− ∫_{−∞}^{0} xf_{Y} (x)dx]
= ∫_{0}^{∞} P {Y > y}dy − ∫_{0}^{∞} P {Y < −y}dy  (11)

5. Use the result that for a nonnegative random variable Y

E[Y] = ∫_{0}^{∞} P {Y > t}dt

to show for a nonnegative random variable X

E[X^{n}] = ∫_{0}^{∞} nx^{n−1}P {X > x}dx  (12)

By definition,

E[X^{n}] = ∫_{0}^{∞} P {X^{n} > t}dt  (13)

8. Given that P {0 ≤ X ≤ c} = 1, show

V ar(X) ≤ \frac{c^{2}}{4}  (14)

To illustrate (14), will use the following results.

E[X^{2}] = E[X · X]
≤ E[c · X]
= cE[X]  (15)

V ar(X) = E[X^{2}] − (E[x])^{2}
≤ cE[X] − (E[x])^{2}
= c^{2} \frac{E[X]}{c} − c^{2} (\frac{E[X]}{c})^{2}

= c^{2} (\frac{E[X]}{c} − (\frac{E[X]}{c})^{2})

= c^{2} (\frac{E[X]}{c} [1 − \frac{E[X]}{c}])  (16)

Let α = \frac{E[X]}{c} in (16) to obtain,

V ar(X) ≤ c^{2}[α(1 − α)]  (17)

The maximum value for α = \frac{1}{2}, solved by finding vertex of f(x) = x − x^{2}.

V ar(X) ≤ \frac{c^{2}}{4}  (18)

3. Given,

f(x) = \{ \begin{matrix}C(2x − x^{3}) & 0 < x < \frac{5}{2} \\ 0 & else \end{matrix}  (19)

The above function (19) is not a p.d.f, because f(x) must be nonnegative within domain. Determine C.

By definition,

∫_{−∞}^{∞} f(x)dx = 1

Therefore,

∫0^{\frac{5}{2}} C(2x − x^{3})dx = 1
C [x^{2} − \frac{x^{4}}{4} ] |0^{\frac{5}{2}} = 1
C (−\frac{225}{64}) = 1
C = −\frac{64}{225}  (20)

f(x) = \{ \begin{matrix}−\frac{64}{225}(2x − x^{2}) & 0 < x < \frac{5}{2} \\ 0 & else \end{matrix}  (21)

Given,

f(x) = \{ \begin{matrix}C(2x − x^{2}) & 0 < x < \frac{5}{2} \\ 0 & else \end{matrix}  (22)

The above function (22) is not a p.d.f, because f(x) must be nonnegative within domain. Determine C.

By definition,

∫_{−∞}^{∞} f(x)dx = 1

Therefore,

∫0^{\frac{5}{2}} C(2x − x^{2})dx = 1
C [x^{2} − \frac{x^{3}}{3} ] |0^{\frac{5}{2}} = 1
C (\frac{25}{24}) = 1
C = \frac{24}{25}  (23)

f(x) = \{ \begin{matrix}\frac{24}{25}(2x − x^{2}) & 0 < x < \frac{5}{2} \\ 0 & else \end{matrix}  (24)

f(x) = \{ \begin{matrix}\frac{10}{x^{2}} & x > 10 \\ 0 & x ≤ 10 \end{matrix}  (25)

a. Determine the following

P (x > 20) = ∫_{20}^{∞} \frac{10}{x^{2}}dx
= 1 − ∫_{0}^{20} \frac{10}{x^{2}}dx
= 1 − ∫_{10}^{20} \frac{10}{x^{2}}dx
= 1 − [−\frac{10}{x}] |_{10}^{20}
= 1 − \frac{1}{2}
= \frac{1}{2}  (26)

b. What is the c.d.f of f(x)? by defintion,

\frac{d}{dx}F (x) = f(x)  (27)