Gaussian Distribution Introductory Level-quiz PDF

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Gaussian Distribution assignment.Questions to test understanding of Normal Distibutions.Intermediate level maths knowledge required...

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Quiz-4 answers and solutions Coursera. Stochastic Processes December 30, 2020 1. Choose the functions K (t, s), which can be covariance functions of some stochastic processes defined for t ∈ [0, 1]. Options: a) K (t, s) = 1 − max{t, s} c) K (t, s) = 1 + (t − s)3

b) K (t, s) = max{t, s} − min{t, s} d) c > 0

Answer: a) K (t, s) = 1 − max{t, s}, d) c > 0 Solution: a) Clearly, this function is symmetric, since K (t, s) = 1 − max{t, s} = 1 − max{s, t} = K (s, t). Let us show that it is also positive semi-definite. Indeed, define gt (x) := I{x ∈ [t, 1]}. Then Z∞ 1 − max{t, s} = gs (x)gt (x) dx, 0

from which n n X X

uj uk K (tj , tk ) =

n n X X

uj uk

j=1 k=1

j=1 k=1

=

Z∞X n X n 0

Z∞

gtj (x)gtk (x) dx

0

uj uk gtj (x)gtk (x) dx

j=1 k=1

  ! Z∞ X n n X   = uj gtj (x) uj gtk (x) dx j=1

0

k=1

 2 Z∞ X n  = uj gtj (x) dx ≥ 0. j=1

0

Thus, K (t, s) = 1−max{t, s} can be a covariance function of some process defined for t ∈ [0, 1].

b) This function cannot be a covariance function of any process. Assume the converse: let there exist some process Xt such that its covariance function K (t, s) = max{t, s} − min{t, s}. Then Var Xt = K (t, t) = max{t, t} − min{t, t} = t − t = 0, 1

i.e., Xt = const for all t ∈ [0, 1]. However, cov(Xt , Xs ) = max{t, s} − min{t, s} = 6 const

∀s 6= t,

(s, t) ∈ [0, 1]2 ,

which leads to the contradiction. c) Since

K (t, s) = 1 + (t − s)3 6= 1 + (s − t)3 = K (s, t),

the symmetry property is violated, and this function cannot be a covariance function of any process. d) It can be seen that K (t, s) = c = K (s, t) and n n X X

uj uk K (tj , tk ) =

n n X X j=1 k=1

j=1 k=1



uj uk c = c 

n X j=1

2

uj  ≥ 0,

meaning that this function is symmetric and positive semi-definite. Thus, it can be a covariance function of some process. A corresponding example is given by Xt = ξ, where ξ is a random variable such that Var ξ = 1. 2. Which of the following processes are Gaussian? (In the answers below Wt is a Brownian motion). Options: a) Xt = Wt2 b) Xt = aξt−1 + ξt , c) Xt = W1−t − W1 , d) Xt = Wt + ηξ,

ξi ∼ i.i.d. N (0, 1) ∀i = 0, 1, . . . , t,

t = 1, 2, . . .

t ∈ [0, 1]

( −1, 1/2 ξ ∼ N (0, 1), η = , 1, 1/2

Wt , ξ, η are indepen-

dent for all t ≥ 0 Answer: b), c), d) Options: The process Xt is Gaussian if and only if for any t1 , . . . , tn : n P λk Xtk ∼ N , ∀~λ. (Xt1 , . . . , Xtn ) is a Gaussian vector, that is, k=1

a) Since P{W t2 < 0} = 0, Wt2 is not normally distributed, and this process is not Gaussian. b) A finite dimensional distribution for this process has the form (Xt1 , Xt2 , . . . , Xtn ) =

(aξt1 −1 + ξt1 , aξt2 −1 + ξt2 , . . . , aξtn −1 + ξtn ).

Clearly, any linear combination of finite dimensional distributions of Xt can be represented as a sum of i.i.d. normally distributed random variables: for instance, if ti are such that ti − 1 = ti−1 ∀i = 1, 2, . . . , n n X

λk Xtk

=

λ1 (aξt1 −1 + ξt1 ) + λ2 (aξt2 −1 + ξt2 ) + · · · + λn (aξtn −1 + ξtn )

=

λ1 (aξt1 −1 + ξt1 ) + λ2 (aξt1 + ξt2 ) + · · · + λn (aξtn−1 + ξtn )

k=1

=

λ1 aξt1 −1 + (λ1 + aλ2 )ξt1 + (λ2 + aλ3 )ξt1 + · · · + λn ξtn ∼ N . 2

c) This process is Gaussian as n X

λk Xtk

=

k=1

=

λ1 W1−t1 − λ1 W1 + · · · + λn W1−tn − λn W1 n X k=1

λk W1−tk − W1

n X k=1

λk ∼ N

as a linear combination of the components of the finite dimensional distribution (W1−t1 , W1−t2 , . . . W1−tn , W1 ) of a Brownian motion which is a Gaussian process. d) Let us show that ηξ ∼ N (0, 1). Indeed, P{ηξ ≤ x}

P{−ξ ≤ x|η = −1}P{η = −1} + P{ξ ≤ x|η = 1}P{η = 1} 1 1 (1 − P{ξ > x} + P{ξ ≤ x}) = (1 − 1 + P{ξ ≤ x} + P{ξ ≤ x}) = 2 2 = P{ξ ≤ x}.

=

Thus n X

λk Xtk =

k=1

n X

λk (Wtk + ηξ) =

k=1

n X k=1

˜λk (Wt − Wt ) + ηξ k k−1

n X k=1

λk ∼ N

as a sum of i.i.d. normally distributed random variables. 3. Which of the following processes is a Brownian motion? (In the answers below Wt is a Brownian motion). Options: a) Xt = Wt2 b) Xt = aξt−1 + ξt , c) Xt = W1−t − W1 , d) Xt = Wt + ηξ, dent for all t ≥ 0

ξi ∼ i.i.d. N (0, 1) ∀i = 0, 1, . . . , t,

t = 1, 2, . . .

t ∈ [0, 1]

( −1, 1/2 ξ ∼ N (0, 1), η = , 1, 1/2

Answer: c) Xt = W1−t − W1 ,

Wt , ξ, η are indepen-

t ∈ [0, 1]

Solution: Since Brownian motion is a Gaussian process Wt such that EWt = 0 and cov(Wt , Ws ) = min{t, s}, we should only check the last two properties for the processes b)-d). For b) E[Xt ] = E[aξt−1 + ξt ] = aE[ξt−1 ] + E[ξt ] = 0, but cov(Xt , Xs ) = cov(aξt−1 +ξt , aξs−1 +ξs ) = (1+a2 )I{t = s}+aI{|t−s| = 1}, thus, it is not a Brownian motion. For c) E[W1−t − W1 ] = E[W1−t ] − E[W1 ] = 0 3

and cov(W1−t − W1 , W1−s − W1 ) =

cov(W1−t , W1−s ) − cov(W1 , W1−t )

− cov(W1 , W1−s ) + cov(W1 , W1 )

min{1 − t, 1 − s} − (1 − t) − (1 − s) + 1

=

t + s − max{t, s} = min{t, s}.

= Therefore, Xt is Brownian motion. For d)

E[Wt + ηξ] = E[Wt ] + E[η]E [ξ] = 0, but cov(Wt + ηξ , Ws + ηξ)

=

cov(Wt , Ws ) + cov(Wt , ηξ) + cov(Ws , ηξ ) + cov(ηξ, ηξ )

=

min{t, s} + 0 + 0 + E[(ηξ)2 ] − (E[ηξ ])2

ηξ∼N (0,1)

=

min{t, s} + 1 − 0

min{t, s} + 1 6= min{t, s}.

= So, Xt is not a Brownian motion.

4. Find the covariance function of the process Xt = Wt − tW1 , where Wt is a Brownian motion.

t ∈ [0, 1],

Answer: min{t, s} − ts

Solution:

cov(Xt , Xs ) = = =

cov(Wt − tW1 , Ws − sW1 )

cov(Wt , Ws ) − t cov(W1 , Ws ) − s cov(Wt , W1 ) + ts cov(W1 , W1 )

min{t, s} − ts − st + ts = min{t, s} − ts.

5. Find the covariance function of the process Xt = ξWt , independent of Wt ∀t, t ≥ 0.

ξ ∼ N (1, 1) is

Answer: 2 min{t, s}.

Solution:

cov(ξWt , ξWs )

= = t>s

=

= =

E[ξ 2 Wt Ws ] − E[ξWt ]E[ξWs ]

Eξ 2 E[Wt Ws ] − (E[ξ])2 EWt EWs

(1 + 12 )E[(Wt − Ws + Ws )Ws ] − 02 · 0 · 0

2(E[(Wt − Ws )(Ws − W0 )] + E[Ws2 ]) 2s = 2 min{t, s}.

6. Harry is playing some game. At each given time moment t the number of points he gets is determined by the value of the process Xt = σWt , that is, he earns points if Xt > 0 and loses if Xt < 0. Every hour (at time moments t = 1, 2, . . .) the results are automatically recorded, and if the sum of 3 results in a row is greater than 10, Harry gets a special prize. Find the probability that Harry gets this prize after 3 hours of playing (in 4

the answers below Φ is the distribution function of the standard normal distribution).   10 Answer: 1 − Φ √ σ 14

Solution: We need to find the probability that P{X1 + X2 + X3 ≥ 10}

= =

P{σ(W1 + W2 + W3 ) ≥ 10}

1 − P{W1 + W2 + W3 ≤ 10/σ }.

As E[W1 + W2 + W3 ] = E[W1 ] + E[W2 ] + E[W3 ] = 0 and Var(W1 + W2 + W3 ) =

Var W1 + Var W2 + Var W3 + 2 cov(W1 , W2 ) +2 cov(W2 , W3 ) + 2 cov(W1 , W3 )

=

1 + 2 + 3 + 1 · 2 + 2 · 2 + 1 · 2 = 14,

1 − P{W1 + W2 + W3 ≤ 10/σ} = 1 − Φ



10 √ σ 14



.

7. Choose the correct statements about the process Xt = σWt , σ > 0: Options: a) cov(Xt , Xs ) = min{t, s }

b) the Kolmogorov continuity theorem E[|Xt − Xs |α ] ≤ c|t − s|1+β holds with c = 3, α = 4, β = 1 c) Xt is not stationary d) Xt is a Gaussian process Answer: Xt is not stationary and is a Gaussian process. Solution: Since Xt is a Brownian motion multiplied by a positive constant, it is also a Gaussian process. Now, cov(Xt , Xs ) = σ 2 cov(Wt , Ws ) = σ 2 min{t, s} and since the covariance function depends not on the difference, but on the time moments t and s, Xt is not stationary. Finally, E[|Xt − Xs |α ] = E[|σ(Wt − Ws )|α ] = σ α E[|Wt − Ws |α ], therefore, the condition of the Kolmogorov continuity theorem E[|Xt − Xs |α ] ≤ c|t − s|1+β holds with α = 4, β = 1 and c = 3σ 4 .

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