April 05, ques 3

[MindFull]

For anyone that's done this past paper before, can i have an explanation to how this ques is answered by the examiners really not sure what's happening here... Really confused about the E(Y2k) = 0 , and what P(Y2k-1 = 1 ) is.

Any help would be appreciated.

Here's the ques.
Let Y1, Y3, Y5,,, be a sequence of independent and identically distributed random
variables with
2 1 2 1
1
=1 = = 1 = , = 0, 1, 2,...
P Y k P Y k 2 k
and define Y2k = Y2k 1 /Y2k 1 for k = 1, 2, .
(i) Show that Yk : k =1, 2,... is a sequence of independent and identically
distributed random variables.

 

[Robert Chadburn]

You are given the probability distributions of all the values Y1, Y3, Y5 etc. The value can be either +1 or -1 with equal probability. And we are told that the values are independent of each other.

So you need to find the probability distribution of the values Y2, Y4, Y6, etc and also establish whether these are independent of themselves and of the other values.

Start with one example, say Y2. It is calculated as Y3/Y1. So its values depend on Y3 and Y1. Y2=1 if Y3 and Y1 have the same sign. You should calculate the probability of this and find that it is a half. Then also Y2=-1 if Y1 and Y3 are of different sign - again calculate the probability of this and find it is again one half.

So you now have the prob distribution of Y2 (and all the other Y2k): it takes the value 1 or -1 with equal probability, which is the same distribution as all the other values, as required.

You also have to test for independence, but the question gives you a method to use for this. I would take, for example, E(Y4xY5), and check that it was equal to E(Y4)xE(Y5) (ie equals zero). To do this you would have to calculate the E(XY) formula properly, ie:

E(XY) = 1xPr([X=1 and Y=1] or [X=-1 and Y=-1]) + -1xPr([X=-1 and Y=1] or [X=1 and Y=-1])

in which you would have to use conditional probabilities, such as:

Pr(X=1 and Y=1) = Pr(X=1) x Pr(Y=1 given X=1) etc.

This is what I would have done, anyway. Hope it helps a bit.
Good luck!

 

[jm_kinuthia]

Hi

Kindly help me understand the following:

1. The answer suggests that pairewise independence was implied. How different would it be if mutual independence was required ?

2. Also, how is E(Y2k,Y2k+1) calculated ? I can't understand the solution in the report.

Thanks.

 

[Hemant Rupani]

Hi

Kindly help me understand the following:

1. The answer suggests that pairewise independence was implied. How different would it be if mutual independence was required ?

2. Also, how is E(Y2k,Y2k+1) calculated ? I can't understand the solution in the report.

Thanks.

1. mutual independence may differ if you were finding independence of more than 2 RVs/events/processes.

2. I guess you are asking for \( E(Y_{2k}Y_{2k+1}) \)...... as it depends on \( Y_{2k-1} \), it is equal to \( P(Y_{2k-1}=-1)\times E(Y_{2k}Y_{2k+1}|Y_{2k-1}=-1) + P(Y_{2k-1}=1)\times E(Y_{2k}Y_{2k+1}|Y_{2k-1}=1) \)

Now, \( P(Y_{2k-1}=-1)=P(Y_{2k-1}=-1)=\frac{1}{2} \) and
\( E(Y_{2k}Y_{2k+1}|Y_{2k-1}=-1)=-1.........~as~for~Y_{2k+1}=-1, Y_{2k}=1~and~for~Y_{2k+1}=1, Y_{2k}=-1 \) and
\( E(Y_{2k}Y_{2k+1}|Y_{2k-1}=1)=1.........~as~for~Y_{2k+1}=-1, Y_{2k}=-1~and~for~Y_{2k+1}=1, Y_{2k}=1 \)