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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, . . . ,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. (c) Are X and Y independent? Why?

Respuesta :

Answer:

a) [tex] P_X(k) = 0.2 [/tex] for any k in {1,2,3,4,5}

[tex] P_Y(1) = 0.45666

P_Y(2) = 0.25666

P_Y(3) = 0.15666

P_Y(4) = 0.09

P_Y(5) = 0.05

b)

  • [tex] P(X=j| \, Y=1) = \frac{1}{5P(Y=1)} = \frac{1}{5jP(Y=1)} = 0.4379/j [/tex] for any j n {1,2,3,4,5}
  • [tex] P(X=j| \, Y=2) = \frac{1}{5jP(Y=2)} = 0.7792/j [/tex] for j > 1
  • [tex] P(X=j| \, Y=3) = \frac{1}{5jP(Y=3)} = 1.2766/j [/tex] for j > 2
  • [tex] P(X=j| \, Y=4) = \frac{1}{5jP(Y=4)} = 0.555 [/tex] if j = 4 or 0.444 if j = 5
  • [tex] P(X=5| \, Y=5) = 1 [/tex]

c) No

Step-by-step explanation:

For X, we have 1 chance out of five to get any of the five numbers, so [tex] P_X(k) = 1/5 = 0.4 \, [/tex] 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.

  • [tex]P_Y(1) = P(Y=1) = P(Y=1 | \, X = 1) * P(X=1) + \\P(Y=1| \, X=2) * P(X=2) + P(Y=1 | \, X=3) * P(X=3) + \\P(Y=1 | \, X=4)*P(X=4) + P(Y=1 | X=5)*P(X=5) = \\01*0.2 + 0.5*0.2+1/3*0.2+0.25*0.2+0.2*0.2 = 0.45666[/tex]
  • [tex]P_Y(2) = P(Y=2) = P(Y=2 | \, X = 2) * P(X=2) + \\P(Y=2| \, X=3) * P(X=3) + P(Y=2 | \, X=4) * P(X=4) + \\P(Y=2 | \, X=5)*P(X=5) = 0.5*0.2+1/3*0.2+0.25*0.2+0.2*0.2 = 0.25666[/tex]
  • [tex] P_Y(3) = P(Y=3) = P(Y=3| \, X=3) * P(X=3) + \\P(Y=3 | \, X=4) * P(X=4) + P(Y=3 | \, X=5)*P(X=5)\\ = 1/3*0.2+0.25*0.2+0.2*0.2 = 0.15666[/tex]
  • [tex]P_Y(4) = P(Y=4) = P(Y=4 | \, X=4) * P(X=4) + \\P(Y=4 | \, X=5)*P(X=5) = 0.25*0.2+0.2*0.2 = 0.09[/tex]
  • [tex]P_Y(5) = P(Y=5) = P(Y =5 | \, X=5) = 0.2*0.2 = 0.04[/tex]

b) In order to solve this item, we should use Bayes Formula. Note that for any i, j we have

[tex] P(X = j | \, Y = i) = \frac{P(Y = i | \, X=j) * P(X=j)}{P(Y=i)} [/tex]

This probability is always 0 if j is smaller than i. If j is bigger or equal than i, then P(Y = i |X=j) is 1/j. Also P(X=j) = 0.2. Therefore, for i ≤ j, we have

[tex] P(X = j | \, Y = i) = \frac{P(Y = i | \, X=j) * P(X=j)}{P(Y=i)} = \frac{0.2/j}{P(Y=i)} = \frac{1}{5jP(Y=i)}[/tex]

Since we calculated the values of P(Y=i), we alredy know this values

  • [tex] P(X=j| \, Y=1) = \frac{1}{5P(Y=1)} = \frac{1}{5jP(Y=1)} = 0.4379/j [/tex] for any j n {1,2,3,4,5}
  • [tex] P(X=j| \, Y=2) = \frac{1}{5jP(Y=2)} = 0.7792/j [/tex] for j > 1
  • [tex] P(X=j| \, Y=3) = \frac{1}{5jP(Y=3)} = 1.2766/j [/tex] for j > 2
  • [tex] P(X=j| \, Y=4) = \frac{1}{5jP(Y=4)} = 0.555 [/tex] if j = 4 or 0.444 if j = 5
  • [tex] P(X=5| \, Y=5) = 1 [/tex]

c) X and Y arent independent because if they were, we should have that P(X=j) shoudnt change if we condition anything from Y, but P(X=j) = 0.2, and we obtained results different from 0.2 when we conditioned by different values of Y in the previous item.

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