Statistics - Dependence/independence of variables

[peripatein]

Hi,

Homework Statement

The sample space of the following problem is defined thus: all the possible permutations of {1,2,3} including {1,1,1}, {2,2,2}, {3,3,3}. Suppose all results are equally probable. Let X_i denote the value of the ith coordinate, where i=1,2,3.
I am asked to determine whether, for any i≠j, X_i and X_j are independent and whether {X₁,X₂,X₃} are independent.

Homework Equations

The Attempt at a Solution

The sample space is 9, I believe. And the probability of each set is 1/9 (even probability). What perplexes me is how to determine whether X_i and X_j are dependent/independent. A hint indicates that I may prove it for any i and j I choose. Suppose I choose i=1 and j=2. Am I now to check whether P(X₁ \cap X₂) / P(X₂) is equal to P(X₁)?

 

[I like Serena]

From wikipedia:

Two events A and B are independent if and only if their joint probability equals the product of their probabilities:
$$P(A \cap B)=P(A)P(B)$$​

So yes, you're on the right track. ;)

 

[Ray Vickson]

Hi,

Homework Statement

The sample space of the following problem is defined thus: all the possible permutations of {1,2,3} including {1,1,1}, {2,2,2}, {3,3,3}. Suppose all results are equally probable. Let X_i denote the value of the ith coordinate, where i=1,2,3.
I am asked to determine whether, for any i≠j, X_i and X_j are independent and whether {X₁,X₂,X₃} are independent.

Homework Equations

The Attempt at a Solution

The sample space is 9, I believe. And the probability of each set is 1/9 (even probability). What perplexes me is how to determine whether X_i and X_j are dependent/independent. A hint indicates that I may prove it for any i and j I choose. Suppose I choose i=1 and j=2. Am I now to check whether P(X₁ \cap X₂) / P(X₂) is equal to P(X₁)?

Don't forget that $\{X_1, X_2,X_3\}$ independent also requires that
P(X_1=i \:\& \: X_2=j \:\& \: X_3=k) = P(X_1=i) P(X_2=j) P(X_3=k) for all i, j and k.

 

[Ray Vickson]

Hi,

Homework Statement

The sample space of the following problem is defined thus: all the possible permutations of {1,2,3} including {1,1,1}, {2,2,2}, {3,3,3}. Suppose all results are equally probable. Let X_i denote the value of the ith coordinate, where i=1,2,3.
I am asked to determine whether, for any i≠j, X_i and X_j are independent and whether {X₁,X₂,X₃} are independent.

Homework Equations

The Attempt at a Solution

The sample space is 9, I believe. And the probability of each set is 1/9 (even probability). What perplexes me is how to determine whether X_i and X_j are dependent/independent. A hint indicates that I may prove it for any i and j I choose. Suppose I choose i=1 and j=2. Am I now to check whether P(X₁ \cap X₂) / P(X₂) is equal to P(X₁)?

Don't forget that $\{X_1, X_2,X_3\}$ independent also requires that
P(X_1=i \:\& \: X_2=j \:\& \: X_3=k) = P(X_1=i) P(X_2=j) P(X_3=k) for all i, j and k.

 

[peripatein]

But what is P(X1)? What does it stand for in this case? Is it the probability that the first coordinate would be 1,2 or 3? Which makes no sense, as wouldn't that be 1?
This is the part I am not sure I am grasping in the question. Could someone please clarify?

 

[Ray Vickson]

But what is P(X1)? What does it stand for in this case? Is it the probability that the first coordinate would be 1,2 or 3? Which makes no sense, as wouldn't that be 1?
This is the part I am not sure I am grasping in the question. Could someone please clarify?

You are confusing yourself by using bad notation: $P(X_1)$ has no meaning! The meaningful statements about $X_1$ are of the form $P(X_1=1) = ?$ (you fill it in), as well as $P(X_1 = 2) = ?$ and $P(X_1 = 3) = ?$ In general, you must determine the values of the probabilities $P(X_i = j)$ for i = 1,2,3 and j = 1,2,3.

 

[peripatein]

Alright, so suppose I choose i=1 and j=2, would that be the probability P(X₁ \capX₂ = 1 | X₃ = 1 (i.e. the probability of X₃ = 1 given that X₁ AND X₂ = 1) + P(X₁ \capX₂ = 2 | X₃ = 2) + P(X₁ \capX₂ = 3 | X₃ = 3) / P(X₂ = 1) \cup P(X₂ = 2) \cup P(X₂ = 3)?

 

[peripatein]

But isn't that P(X1 ∩X2 = 1 | X3 = 1) / P(X2 = 1) + P(X1 ∩X2 = 2 | X3 = 2) / P(X2 = 2) + P(X1 ∩X2 = 3 | X3 = 3) / P(X2 = 3)= 1/3 + 1/3 + 1/3 = 1? What am I doing wrong?

 

[peripatein]

I'd appreciate if any of you were willing to indicate how I am erring in my approach.

 

[I like Serena]

You're confusing me, but it doesn't look right.

Can you perhaps fill in the question marks in Ray's last post?

And can you also find $P(X_1=1 \wedge X_2=1)$?Note that the formula to calculate a probability (assuming equally likely outcomes) is:
$$P(\text{event})=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$$
Can you apply that to find the question marks?

 

[peripatein]

I'd be happy to try.
P(X1=1) = P(X2=1) = P(X3=1) = 3/9 = 1/3.
Now, P(X1=1∧X2=1) = 1/9.
Is that correct?

 

[I like Serena]

I'd be happy to try.
P(X1=1) = P(X2=1) = P(X3=1) = 3/9 = 1/3.
Now, P(X1=1∧X2=1) = 1/9.
Is that correct?

Yes. Good!

As you can see P(X1=1∧X2=1) = P(X1=1) P(X2=1) suggesting that they are independent.
How about P(X1=1∧X2=2)?
It is equal to P(X1=1) P(X2=2)?

Or more generally, how about P(X1=1∧X2=j)?
Is it equal to P(X1=1) P(X2=j)?

And P(X1=i∧X2=j)?

Then, we can go to the next stage.
How about P(X1=1∧X2=1∧X3=1)?
Is it equal to P(X1=1) P(X2=1) P(X3=1)?

 

[peripatein]

I believe P(X1=1∧X2=j) = 1/3.
P(X1=1) P(X2=j) = 1/3 * 1 = 1/3.
Is that correct?

 

[I like Serena]

I believe P(X1=1∧X2=j) = 1/3.
P(X1=1) P(X2=j) = 1/3 * 1 = 1/3.
Is that correct?

Ah, no.
It appears you are summing the probabilities for each j.
But that is not intended.
It's just the probability for one particular j.

For j=1, you already had P(X1=1∧X2=j) = 1/9.
Now we're trying to generalize by saying that P(X1=1∧X2=j) = 1/9 for any j.

 

[peripatein]

Okay, so P(X1=i∧X2=j) = 1/9, and P(X1=i) = P(X1=j) = 3/9 = 1/3.
Is it correct now?

 

[I like Serena]

Yes. So in all cases P(X1=i∧X2=j) = P(X1=i) P(X2=j), which implies that X1 and X2 are independent.

 

[peripatein]

But 1/9 is not equal to 1/27, so the triplet is not independent. Correct?

 

[I like Serena]

Correct.