So, P = T P + F N The proportion of people who are not ill will be denoted by N

1 SPA Homework solutions – week I

Marta Wac lawczyk & Maciej Lisicki

1. COVID test A large group of people is tested for COVID-19 disease. The proportion of ill people that are correctly identified with positive test result equals 0.9. Tests’ specificity (i.e. the percentage of healthy people who are correctly identified as not having some illness) equals 0.95. It is known that 10% of people from the group get positive COVID test result. Determine the true proportion of sick individuals in this group.

Solution Let us denote the proportion of ill people by P (positives) Some of these people will get positive test result T P (true positives), however some of them will receive a false negative result F N (false negative).

So,

P = T P + F N

The proportion of people who are not ill will be denoted by N. Some of them receive negative test result TN (true negative), the rest receive false positive result FP (false pasitive).

N = T N + F P

The number of people in the group will be denoted by L 10% of people in the group gets positive test result, this includes true positives (TP) and false positives (FP)

0.1L = T P + F P

The remaining part of people in the group received negative result, hence 0.9L = T N + F N

We know the test sensitivity (The proportion of ill people that are correctly identified with positive test result) equals 0.9. Hence,

0.9 =T P

P = T P

T P + F N.

The test specificity (the percentage of healthy people who are correctly identified as not having some illness) is 0.95. Hence

0.95 = T N

N = T N

T N + F P We have now 4 equations with 4 unknowns:

0.1L = T P + F P, (1)

0.9L = T N + F N, (2)

0.9 = T P

P = T P

T P + F N, (3)

0.95 = T N

N = T N

T N + F P. (4)

The solution is:

T P = 9

170L (5)

T N = 152

170L (6)

F N = 1

170L (7)

F P = 8

170L (8)

2 Hence the answer is that the true proportion of sick individuals is: (T P + F N )/L = 10/170 ≈ 6%.

2. Stirling’s formula Use the Stirling’s formula to estimate the term:

1 · 3 · 5 · · · (2n + 1) Use the Stirling’s formula to estimate the term:

2 · 4 · 6 · · · (2n) Use the two to calculate

∞

Y

n=1

2n 2n − 1

2n 2n + 1 Solution Calculate first:

2 · 4 · · · 2n = 2ⁿn!

With the Stirling’s formula

2 · 4 · · · 2n = 2ⁿn! ∼ 2ⁿ√ 2πn nⁿ

eⁿ Let us denote

x = 1 · 3 · · · (2n + 1) Note that

(2n + 1)! = 1 · 2 · 3 · · · 2n · (2n + 1) = x · (2 · 4 · · · 2n) Hence,

(2n + 1)! = x · 2ⁿn!

and using the Stirling’s formula we obtain

x = (2n + 1)!

2ⁿn! ∼ (2n + 1)√

2π2n ⁽²ⁿ⁾_e2n²ⁿ 2ⁿ√

2πn ⁿ_enⁿ = (2n + 1)

√2 2ⁿnⁿ eⁿ Observe that

∞

Y

k=1

2n 2n − 1

2n

2n + 1 = lim

n→∞

2 · 4 · · · 2n 1 · 3 · · · (2n − 1)

2 · 4 · · · 2n 1 · 3 · · · (2n + 1) Introducing the previously derived formulas

n

Y

k=1

2k 2k − 1

2k

2k + 1 = 2²ⁿ⁺¹πn ⁿ_e2n²ⁿ

(2n + 1) 2^{2n+1 n}_e2n²ⁿ = π n 2n + 1 In the limit n → ∞ we obtain

lim

n→∞π n

2n + 1 = lim

n→∞

π 2 + _n¹ = π

2

3 Hence,

n

Y

k=1

2n 2n − 1

2n 2n + 1 =π

2

3. Tennis players The Father, trying to motivate his Son to practice her tennis skills, promises him a prize if he wins at least two games in a row with the Father and the Club Master playing according to one of the schemes: Father – Club Master – Father or Club Master – Father – Club Master. Which order should the clever Son choose to maximise the probability of winning the prize if the chance to with with the Father is pf= 0.8, while for the Club Master it is pm= 0.4?

Solution In the first scheme (F-M-F) the probability of winning at least twice in a row is p1= pfpmpf+ (1 − p − f )p_mp_f+ p_fp_m(1 − p_f) (three wins, first lost, last lost). In total

p₁= p_fp_m(2 − p_f).

In the second scheme, the analogous probability is p₂= p_fp_m(2 − p_m) but p_m< p_f, so the easier scheme is Master – Father – Master.

4. Bertrand’s box paradox There are three boxes: a box containing two gold coins, a box containing two silver coins, a box containing one gold coin and one silver coin. We choose one of the boxes at random and withdraw one of the coins from it, which happens to be a gold coin. What is the probability that theother coin from the same box is also gold?

Solution [see also Wiki: Bertrand’s box paradox];

P (both gold| first gold) = P (both gold) P (first gold) =

1 3 1 2

=2 3