Check the solution to "Choose a number X at random from the..."

quiquenobi2v6

Answered question

2021-12-14

Choose a number X at random from the set of numbers

{1, 2, 3, 4, 5}

. Now choose a number at random from the subset no larger than X, that is, from

{1, \dots, X}

. Call this second number Y.
a) Find the joint mass function of X and Y.
b) Find the conditional mass function of X given that Y=i. Do it for i=1, 2, 3, 4, 5.
с) Are X and Y independent? Why?

Answer & Explanation

twineg4

Beginner2021-12-15Added 33 answers

We are given that

X \sim D u n if (1, \dots, 5)

. Observe that

Y \leq X

almost certainly and that also

Y ϵ {1, \dots, 5}

(a)
For

k \leq l

we have that

∣ P (Y = k, X = l) = P (Y = h ∣ X = I) P (X = I) = \frac{1}{l} \cdot \frac{1}{5}

.
I

t k > l

, we have that

P (Y = k, X = l) = 0

(b)
Using the law of the total probability, we have that

P (Y = i) = \sum_{l = i}^{5} P (y = i, x = i) = \sum_{l = i}^{5} \frac{1}{5 l}

Now,for

i \leq

we have that

P (X = i ∣ Y = i) = \frac{P (Y = i, X = i)}{P (Y = i)} = \frac{\frac{1}{5 l}}{\sum_{{i = 1}}^{5} \frac{1}{5 l}}

Y \leq X

, take any

k > l

, for example take

k = 2, l = 1

.
Hence

P (Y = k, X = 1) = 0

. On the other hand, it is pretty obvious that

P (Y = k) > O

and

P (X = 1) > 0 .

Thus

P (Y = k, X = 1) \neq P (Y = k) P (X = I)

so they can not be independent.

Thomas White

Beginner2021-12-16Added 40 answers

For X, we have 1 chance out of five to get any of the five numbers, so

P x (k) = \frac{1}{5} = 0.4

for any integer k between 1 and 5.
For Y,we can use the theorem of total probability, because we can divide Y
in five different cases depending on how it went with the value of X. We can
remove impossible cases (of probability 0), those cases happen when X is
smaller than the value we want to get with Y.

P y (1) = P (Y = 1) = P (Y = 1 ∣ X = 1) \times P (X = 1) +

P (Y = 1 ∣ X = 2) \times P (X = 2) + P (Y = 1 ∣ X = 3) \times P (X = 3) +

P (Y = 1 ∣ X = 4) \times P (X = 4) + P (Y = 1 ∣ X = 5) \cdot P (X = 5) =

0.1 \times 0.2 + 0.5 \times 0.2 + \frac{1}{3} \times 0.2 + 0.25 \times 0.2 + 0.2 \times 0.2 = 0.45666

P y (2) = P (Y = 2) = P (Y = 2 ∣ X = 2) \times P (X = 2) +

P (Y = 2 ∣ X = 3) \times P (X = 3) + P (Y = 2 ∣ X = 4) \times P (X = 4) +

P (Y = 2 ∣ X = 5) \times P (X = 5) = 0.5 \times 0.2 + \frac{1}{3} \times 0.2 \mp 0.25 \times 0.2 \times 0.2 \times 0.2 = 0.25666

P y (3) = P (Y = 3) = P (Y = 3 ∣ X = 3) \times P (X = 3) +

P (Y = 3 ∣ X = 4) \times P (X = 4) + P (Y = 3 ∣ X = 5) \times P (X = 5)

= \frac{1}{3} \times 0.2 + 0.25 \times 0.2 + 0.2 \times 0.2 = 0.15666

P y (4) = P (Y = 4) = P (Y = 4 ∣ X = 4) \times P (X = 4) +

P (Y = 4 ∣ X = 5) \times P (X = 5) = 0.25 \times 0.2 + 0.2 \times 0.2 = 0.09

∣ P y (5) = P (Y =

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