Distributions and Probability

[andrew2786]

Hi, I have a particular problem and if anyone could help me with it, I'd greatly appreciate it! It is as follows:

Two tickets are selected at random from a drum which contains five tickets numbered 1, 1, 2, 2 and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the distribution, mean and variance of
(i) X
(ii) Y.
(iii) Determine the joint distribution of X and Y.
(iv) Find the cov (X,Y)
(v) Find ρ(X,Y) [ where ρ is rho ]

I've thought long and hard about this and just can't do it, please help!

Thanks, Andy

[Rassis]

I will help you with the answers for item (i) and part of item (ii). The remaining will then be easy.

(i) All possible combinations are 1 + 1; 1 + 2; 1 + 3; 2 + 2; 2 + 3. Therefore, X is a distribution containing the sums: 1 + 1 = 2, 1 + 2 = 3, 1 + 3 = 4, 2 + 2 = 4, 2 + 3 = 5. That is 2 with frequency 0.2, 3 with frequency 0.2, 4 with frequency 0.4 and 5 with frequency 0.2. Hence, X[2: 0.2; 3: 0.2; 4: 0.4; 5: 0.2]. Now, the mean is given by: 2 x 0.2 + 3 x 0.2 + 4 x 0.4 + 5 x 0.2 = 3.6. And the variance: [0.2 x (2 - 3.6)^2 + 0.2 x (3 - 3.6)^2 + 0.4 x (4 - 3.6)^2 + 0.2 x (5 - 3.6)^2] / 4 = 0.26.
(ii) All possible combinations are 1 + 1; 1 + 2; 1 + 3; 2 + 2; 2 + 3. Therefore, Y is a distribution containing the maximums: 1 & 1 = 1, 1 & 2 = 2, 1 & 3 = 3, 2 & 2 = 2, 2 & 3 = 3, that is 1 with frequency 0.2, 2 with frequency 0.4, 3 with frequency 0.4. Hence, Y[1: 0.2; 2: 0.4; 3: 0.4]. The rest is up to you.

Rui

[andrew2786]

As there are two lots of 1 and 2 (1, 1, 2, 2,) wouldn't that change a lot of what you've said as you could draw 1+2 more than just the once as you could draw the the first 1 ticket and the first 2 ticket and then repeat again but draw the other 1 and the other 2...? I'm pretty sure this would make a difference.

[Rassis]

You are absolutely right. I am sorry for not having noticed it.
You have to account for the probabilities of all possible outcomes:

1 in the first draw and 1 in the second draw = 2/5 x 1/4 = 0.1
1 in the first draw and 2 in the second draw = 2/5 x 2/4 = 0.2
1 in the first draw and 3 in the second draw = 2/5 x 1/4 = 0.1
2 in the first draw and 1 in the second draw = 2/5 x 2/4 = 0.2
2 in the first draw and 2 in the second draw = 2/5 x 1/4 = 0.1
2 in the first draw and 3 in the second draw = 2/5 x 1/4 = 0.1
3 in the first draw and 1 in the second draw = 1/5 x 2/4 = 0.1
3 in the first draw and 2 in the second draw = 1/5 x 2/4 = 0.1
Total = 1.0

Sum the numbers of all the possible outcomes:

1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
2 + 1 = 3
2 + 2 = 4
2 + 3 = 5
3 + 1 = 4
3 + 2 = 5

Now sum the frequencies from above for each one of the possible outcomes:

2: 0.1
3: 0.2 + 0.2 = 0.4
4: 0.1 + 0.1 + 0.1 = 0.3
5: 0.1 + 0.1 = 0.2
Total = 1.0

Therefore:
X[2 x 0.1; 3 x 0.4; 4 x 0.3; 5 x 0.2]

Following the same rationale, you find:
Y[1 x 0.1: 2 x 0.4; 3 x 0.2; 4 x 0.1; 6 x 0.2]

You know already how to proceed from here.

[andrew2786]

Sorry to correct you again, but for the Y distribution, I've worked it out to be:

1 & 1 - maximum - 1
1 & 2 - maximum - 2
1 & 3 - maximum - 3
2 & 1 - maximum - 2
2 & 2 - maximum - 2
2 & 3 - maximum - 3
3 & 1 - maximum - 3
3 & 2 - maximum - 3

& using what you worked out for X, to get the prob. distribution of Y:
p(1) = 0.1
p(2) = 0.2 + 0.2 + 0.1 = 0.5
p(3) = 0.1 + 0.1 + 0.1 + 0.1 = 0.4

So,

Y | 1 2 3
P(Y=y) | 0.1 0.5 0.4


Yes?

[Rassis]

Andy,

I was wrong and your result is right. I used mistakenly both rules – the sum and the maximum for set Y. I am sorry one more time. That was the result of trying to be productive until two o’clock in the morning! I am happy that you noticed it. It means that you got the logic and you will never forget it.

Thank you for your patience and persistence. I hope you’ll succeed your studies.

Rui