So wghats ur name

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University of New South Wales

MATH 2901 Assignment



Question 1:Given that there are n number of random variables, the proba-bilityP (X ≤ x) = P (min(X 1.....X n) ≤ x) suggests that there is atleast one X i that is smaller than x. Also, note that the prob-

ability that at least ONE X i is smaller than x is equivalent to1− P (X i ≥ x).Since X i is identically distributed, the probability of all X i isgreater than x is simply (1 − F X (x))n.Hence the probability that at least ONE X i is smaller than x isthen1− (1 − F X (x))n.

b,F X (x) =

x0

f X (t) dt

substituting the given density function above yield:

F X (x) =

x0

f X (t) dt

= 1

β

x0

t

β e−tβ dt

using integration by parts, we get

1



= −1

β

te

−tβ

x0− x0

e−tβ dt

= −1

β xe−xβ − − βe

−tβ

x

0=

1

β

−xe

−xβ +

−βe

−xβ + β

= 1− e

−xβ

x

β + 1

c,

As the lights are connected are in series, if one breaks, all of thelight bulbs will also break. So to find the expected life time of all the bulbs, we just need to find the time taken for the firstlight bulb fails. Thus taking what we have found in part (a)we let T = min (X 1, ...., X n)Then

F T (x) = 1

− 1

−1

−e−xβ 1 +

x

β

n

= 1−

e−xβ

x

β + 1

n

d,Let Y =

√ nT .

Therefore:

F Y (y) = P (Y ≤ y)= P (

√ nT ≤ y)(using the subsitituion)

= P (T ≤ y√ n

)

∴ F y(y) = 1 −

e−y√ nβ

1 +

y√ nβ

n

So as we take n →∞



limn→∞

F Y (y) = limn→∞ 1−

e

−y√ nβ

1 +

y√ nβ

n

= 1− limn

→∞

eln(e

−y√ nβ (1+ y√

nβ))n

by evaulating inside the limit of the ln function

limn→∞ ln

e

−y√ nβ

1 +

y√ nβ

n

= limn→∞n

−y√ nβ

+ n ln

1 +

y√ nβ

From here, by discussing with a fellow peer S.zhu, he pointedout that I should consider the Taylor expansion of ln(1 + x)

limn→∞ ln

1 +

y

√ nβ

=

y

√ nβ − 1

2

y

√ nβ 2

+

1

3 y

√ nβ 3

− .....

Hence

limn→∞n

−y√

nβ +

y√ nβ −

1

2

y√ nβ

2

+ 1

3

y√

nβ

3

− .....

= lim

n→∞

−1

2

y2

β 2 +

1

3

y3

β 3√

n− .

= −y2

2β 2

So limn→∞ 1−

e−y√ nβ

1 + y√

nβ

n= 1 − e

−y2

2β2

e,

From part(d), we have shown that F Y (y) = 1−e−y2

2β2 To computethe expected value, we need the density function,

so f Y (y) = yβ 2

e−y

2

2β2 The density function resembles the Normaldistribution, so:

E (Y ) =

∞0

t2

β 2e−t2

2β2 dt

=

√ 2π

β

∞0

t2

β 2e−t2

2β2 dt



Also, ∞−∞

t2

β 2e−t2

2β2 dt is simply the second moment of the normaldistribution. According to https://mazeofamazement.wordpress.com/2010/07/03/littlebit-more-gaussian/, the second moment can be calculated byE [X ] = σ2 + µ2

E (Y ) =

√ 2π

2β

∞−∞

t2

β 2e−t2

2β2 dt(since the integrand is an even function)

=

√ 2π

2β (02 + β 2)

=

√ 2π

2 β

Hence E (T ) = E (Y )√ n

= 250√ 2π. The answer is so much lowerwhen n=100 as we expect the first light bulb in the circuit tofail is far less than the life of a single light bulb.

Question 2a,Using the fact that z0

pu p−1 du = z p

E (z p) = ∞

0

z pf Z (u)du

=

∞0

z0

pu p−1f Z (u)dudz

by reversing the order of integration and taking horizontal strips,



∞0

∞u

pu p−1f Z (u)dzdu = p

∞o

u p−1

F Z (u)∞u

= p ∞

0

u p−1(1− F Z (u))du

as required

b,l(m) is defined as E (|X − m|). By considering P=1

l(m) =

∞0

(1− F H (u))du

=

∞0

1− P (|X −m| ≤ u)du

= ∞0

1− P (−u ≤ X − m ≤ u)du

=

∞o

1− P (m− u ≤ X ≤ m + u)du

=

∞0

1− F (m + u) + F (m− u)du

Now, by differentiating both sides and using Liebniz Rule, we

get the following expression

d(l(m)

dm =

d

dm

∞0

1− F (m + u) + F (m− u)du

We now apply Liebniz Rule here:



d(l(m)

dm =

∞0

∂

∂m(1 − F (m + u) + F (m− u)du)

= ∞

0

f (m− u)− f (m + u)du

= −F (m + u)− F (m− u)∞0

= −F (m + ∞)− F (m−∞)− (−F (m + 0) − F (m− 0))

= −1 + 2F (m).

Now to find to find the ’stationary points of d(l(m)dm

, we set it to

equal zero. Therefore, we get the following

−1 + 2F (m) = 0

F (m) = 1

2

m = F −1(1

2)

As l(m) is increasing, the stationary point we have found is aminimum. Hence m∗ = F −1 1

2 occurs at the median of the dis-tribution.

Question 3aTo find the maximum Likelihood (mle) of the of α we considerusing the natural logarithm function. This will allow the differ-entiation easier;



lnL(k, α|x) =ni=1

ln

αkα

xα+1i

= nln

(α

) + αnln

(k

)− (α

+ 1)

n

i=1

ln(

xi)

Differentiating the equation above, we get the following:

dln(L(α))

dα =

n

α + nln(k)−

n

i=1

Setting the derivative to equal zero, and re-arranging we get:

n

α =

ni=1

ln(xi)− nln(k)

Introducing α̂ and k̂

α̂

n =

1ni=1 ln(xi)− nln(k)

α̂ =

nni=1 ln(xi)− nln(k)

= nni=1 ln(xi

k )

b,To show that k̂ follows a Pareto distribution, we can find thecumulative distribution function of k̂. From Question 1 (a),



F ̂k(x) = P (min(xi) ≤ X )

= 1− (1− F Xi(x))n

By integrating the density function given within the question,we reach:

F ∗k (x) = 1 −

1−

1− kα

xα

n

= 1

− kαn

xαn

Thus this follows a Pareto distribution with parameters of nα, k

c,The bias of k̂ can be calculated from E (k̂)−k. So if we calculate:

Bias = ∞k x

nαknα

xnα+1 dx− k

= nαknα

∞k

1

xnα dx− k

= nαknα

nα− 1

−1

xnα−1

∞k− k

= nαknα

n1

1

knα−1

− k

= nαk

nα− 1 −k

= k 1

nα − 1

To find the unbiased estimator, we simply just need to do thefollowing calculation:



E (k̂)− k = 0

E (k̂) = k

nαk

nα − 1

= k.

To make the left hand side to equal to k, we need to multiplyE (k̂) by an unbiased estimator. In this case it is the constantnα−1nα

.

d,Let H = min(X 1, X 2, .....X n)By question 1(a), it is clear to see that:

F H (x) = 1 − (1 − F X (x))n

= 1 −

1−

1− kα

xα

n

= 1 −

kα

xα

n

= 1

− kαn

xαn

Hence, it follows the Pareto Distribution with parameters αn, k

Question 4aFrom watching the 19/05/2016 MATH 2901 lecture video,

limn→∞

E

|X n − X |1 + |X n − X |

= lim

n→∞E

I |X n−X |≤

|X n − X |1 + |X n − X |

+ limn→∞E

I

|X n−X |≥

|X n − X |1 + |X n − X |

Now, limn→∞ E

I

|X n−X

|≤

|X n−X |1+|X n−X |

is limn→∞ E

I

|X n−X

|≤

11+|X n−X |



as it is bounded by .

Also, limn→∞ E

I |X n−X |≥

|X n−X |1+|X n−X |

can be re-written as limn→∞ E

I |X n−X |≥

as x

1+x ≤1.

So now we can write:

limn→∞E

|X n − X |1 + |X n −X |

= lim

n→∞E

I

|X n−X |≤

1

1 + |X n −X |

+ lim

n→∞E

I

|X n−X |≥

≤ limn

→∞E I

|X n−X

|≤+ lim

n

→∞P (|X n − X | ≥ )

≤

As is positive and real and our limit is less than it, we can

conclude that limn→∞ E |X n−X |1+|X n−X |

= 0

part(b),Using the hint provided and a late night discussion with L.Wright,

the function f (x) = x1+x is increasing. So;

I |X n−X |>

1 + ≤ |X n −X |

1 + |X n − X |E

I |X n−X |>

1 +

≤ E

|X n − X |1 + |X n − X |

limn

→∞

1 + P (|X n − X | > ) ≤ lim

n

→∞

E I |X n−X |>

1 + = 0(by part(a))

As 1+

is simply a constant, we can conclude that P (|X n −X | > ) =0. This is the definition of convergence in probability,i.e X n

p→ X



BibliographyNA. 2010. Little bit more Gaussian. [ONLINE] Available at:https://mazeofamazement.wordpress.com/2010/07/03/little-bit-

more-gaussian/. [Accessed 22 May 2016].

Joseph Lee Petersen. 2012. Estimating the Parameters of aPareto Distribution. [ONLINE] Available at: http://citeseerx.ist.psu.edu/viewdoc/dow[Accessed 22 May 2016].

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So wghats ur name