# RomanLyapin TheoryAssignment .pdf

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Theoretical Assignment

DeepBayes Summer School 2018 (deepbayes.ru)

Roman Lyapin

April 23, 2018

Exercise 1. The random variable ξ has Poisson distribution with the parameter λ. If ξ = k

we perform k Bernoulli trials with the probability of success p. Let us define the random

variable η as the number of successful outcomes of Bernoulli trials. Prove that η has Poisson

distribution with the parameter pλ.

Solution . By definition η =

Pξ

i=1

{bi = 1} where bi ∼ Ber(p) and ξ ∼ P oi(λ). Then

p(η = k) =

∞

X

p(η = k, ξ = n)

n=0

=

=

∞

X

n=0

∞

X

p(η = k|ξ = n)p(ξ = n)

p(η = k|ξ = n)p(ξ = n)

n=k

As everything is (assumed) independent, conditioned on ξ = n, η ∼ B(n, p) =⇒

n k

p(η = k|ξ = n) =

p (1 − p)n−k

k

ξ is Poisson-distributed, p(ξ = n) =

λn e−λ

n!

=⇒

1

p(η = k) =

=

∞

X

n=k

∞

X

n=k

=

λn e−λ

n!

pk (1 − p)n−k

k!(n − k)!

n!

(λp)k [λ(1 − p)]n−k −λ e−λp

e

k!

(n − k)!

e−λp

∞

(λp)k e−λp X [λ(1 − p)]n−k −λ(1−p)

e

k!

(n

−

k)!

n=k

∞

ˆ n e−λˆ

(λp)k e−λp X (λ)

=

k!

n!

n=0

=

(λp)k e−λp

k!

It follows that η is Poisson-distributed with parameter pλ.

Exercise 2. A strict reviewer needs t1 minutes to check assigned application to DeepBayes

summer school, where t1 has normal distribution with parameters µ1 = 30, σ1 = 10. While a

kind reviewer needs t2 minutes to check an application, where t2 has normal distribution with

parameters µ2 = 20, σ2 = 5. For each application the reviewer is randomly selected with 0.5

probability. Given that the time of review t = 10, calculate the conditional probability that

the application was checked by a kind reviewer.

Solution . It is given that p(r=strict) = p(r=kind) = 0.5, p(t|r = strict) ∼ N (30, 10) and

p(t|r = kind) ∼ N (20, 5).

p(t = 10, r = kind)

p(t = 10)

p(t = 10|r = kind)p(r = kind)

=

p(t = 10|r = kind)p(r = kind) + p(t = 10|r = strict)p(r = strict)

p(t = 10|r = kind)

=

p(t = 10|r = kind) + p(t = 10|r = strict)

p(r = kind|t = 10) =

When calculating normal pdf densities, in both cases

p(r = kind|t = 10) =

=

2

(t−µ)2

2σ 2

1

σkind

1

1

+ σstrict

σkind

1

5

1

1

+

5

10

= 2 =⇒

It follows that the probability the application was reviewed by kind reviewer is 23 .

3

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