Asking for some help with a probability exercise

[squirtle24]

Thanks for any advice. I'd just like to know where I'm going wrong with this.

This is from an exercise in CLRS that I've been mulling over for the last week, but I haven't really figured out where I'm going wrong.

Question: Use indicator random variables to compute the expected value of the sum of n dice.

Before even starting on the problem, we know that the expected value of the sum of n dice would simply be n times the expected value(mean) of one die roll, which is 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = 3.5

Therefore, the expected value of the sum of n dice would be 3.5n.

This makes perfect sense, but I can't seem to figure out how to use indicator random variables to do the same problem. Going by the text, this is what I came up with:

Given a sample space S and event A, an indicator random variable is defined as:
I{A} = { 1 if A occurs, 0 if A does not occur

Let X be the event of a die roll.
Let Xi be the event where the die roll results in an event of i, where 1 <= i <= 6

This gives:
Xi = I{i} = { 1 if i occurs, 0 if i does not occur

By a lemma in the text:
E[X(A)] = Pr{A}

So that gives:
E[Xi] = Pr{i} = 1/6

It follows that:
E[X] = E[X1] + E[X2] + E[X3] + E[X4] + E[X5] + E[X6]

The expected value of a single die roll, then, would be:
E[X] = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6

Obviously, this is not quite there. Somewhere along the way, the outcome value needs to be factored in, but I can't give a reason other than that the fact that I know it should be in there.

I'd appreciate any help. Thanks.

This class is pretty annoying. Homework(textbook questions) is not graded, but exam questions are styled after the textbook questions, and we don't get solutions. Makes it kind of hard to take a test when you don't know if you've been doing the homework correctly or not.

 

[potoba]

"Obviously, this is not quite there. Somewhere along the way, the outcome value needs to be factored in, but I can't give a reason other than that the fact that I know it should be in there. "

I dont know what you meant by that

 

[squirtle24]

Sorry, that was pretty vague. I was talking about the value of the die roll.

So instead of
E[X] = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6

it should be
E[X] = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6

I know it should be this way because I already know what the expected value of a die roll should be. I just don't know how to prove it using this method.

 

[potoba]

The E(X) you expressed by indicator random variable method is the total expectation value of the events (that's why it came out as 1), not the expectation value of the score you earn.
E(X)= sum(f(X)*P(X))

 

[Parasitic]

I haven't done this awhile, but look at Jim Pitman's Probability text...there's a section in there I believe.

Let X = the event that a die roll with side x shows up; then X = sum(Xi), i = 1...6

E(X) = sum(xi P(xi)), i = 1...6, E(X) then = sum(xi, E(Ix)), i = 1..6.
So then E(X) = 1 E(I1) + 2 E(I2) + 3 E(I3)... = 1 (1/6) + 2 (1/6) ...

(based on the identity that E(Ix) = P(x))
Xi is the indictator function for i = x.

 

[potoba]

Originally posted by: squirtle24
Thanks for any advice. I'd just like to know where I'm going wrong with this.

This is from an exercise in CLRS that I've been mulling over for the last week, but I haven't really figured out where I'm going wrong.

Question: Use indicator random variables to compute the expected value of the sum of n dice.

Before even starting on the problem, we know that the expected value of the sum of n dice would simply be n times the expected value(mean) of one die roll, which is 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = 3.5

Therefore, the expected value of the sum of n dice would be 3.5n.

This makes perfect sense, but I can't seem to figure out how to use indicator random variables to do the same problem. Going by the text, this is what I came up with:

Given a sample space S and event A, an indicator random variable is defined as:
I{A} = { 1 if A occurs, 0 if A does not occur

Let X be the event of a die roll.
Let Xi be the event where the die roll results in an event of i, where 1 <= i <= 6

This gives:
Xi = I{i} = { 1 if i occurs, 0 if i does not occur

By a lemma in the text:
E[X(A)] = Pr{A}

So that gives:
E[Xi] = Pr{i} = 1/6

It follows that:
E[X] = E[X1] + E[X2] + E[X3] + E[X4] + E[X5] + E[X6]

The expected value of a single die roll, then, would be:
E[X] = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6

Obviously, this is not quite there. Somewhere along the way, the outcome value needs to be factored in, but I can't give a reason other than that the fact that I know it should be in there.

I'd appreciate any help. Thanks.

This class is pretty annoying. Homework(textbook questions) is not graded, but exam questions are styled after the textbook questions, and we don't get solutions. Makes it kind of hard to take a test when you don't know if you've been doing the homework correctly or not.

i think you either used the lemma wrong or misunderstood the lemma. make sure you understand that E[X(A)] is in the expression "E[X(A)] = Pr{A}"

 

[squirtle24]

I wonder if I've been trying to copy the text example too closely.

I guess I failed to define E[X] correctly.

If X is the random variable associated with the event of a die roll, then
E[X] = sum(i * E[Xi]) from i=1 to i=6
simply because that is the very definition of the expectation of a die roll? I guess I'm going to go with that.

Thank you both for your time and replies.

I included the lemma below. May as well, since I already typed it up.


The lemma is as follows (where X(A) is actually X-sub-A):
Given a sample space S and an event A in the sample space S, let X(A) = I{A}.
Then E[X(A)] = Pr{A}.

The proof was:
E[X(A)] = E[ I{A}] = 1 * Pr{A} + 0* Pr{complement of A} = P{A}

The text had a coin example:
S = {H, T}, with Pr{H} = 1/2 and Pr{T} = 1/2.

Let X(H) be defined as a indicator random variable associated with the coin coming up heads.

X(H) = I{H} = {1 if H occurs, 0 if T occurs

E[X(H)] = E[I{H}] = 1 * Pr{H} + 0 * Pr{T} = 1*1/2 + 0*1/2 = 1/2

I got this result with the die:
E[Xi] = E[I{Xi}] = 1 * Pr{Xi} + 0 * Pr{not Xi} = 1*1/6 + 0*5/6 = 1/6

 

[potoba]

again, E(A) in your case is the probability of the occurrence of the event. It does not reflect the expectation of the score.

 

[squirtle24]

Agreed. As you said, it should be E(X)= sum(f(X)*P(X)). I suppose getting an answer of 1 should have tipped me off.

I think to match up with the text, I would have to write:
E[X] = sum(i * E[Xi]) from i=1 to i=6 = sum(i * P{Xi}) from i=1 to i=6 = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6

 

[potoba]

exactly, use the lemma just to get this = E[I{Xi}] = 1 * Pr{Xi} + 0 * Pr{not Xi} = 1*1/6 + 0*5/6 = 1/6 for each face of the dice
then you have to multiply that with f(X) (=face values of the dice), which goes from 1 to 6 in any dice case, then finally sum them up (this is the definition of expected value)

 

[cohena100]

Here is a valid solution:
For i = 1, 2, ..., 6
Xi be the random variable that counts the number of times that i came out.
Let X be the sum of n dices. X = 1*X1 + 2*X2 + ... + 6*X6
E(Xi) = n/6
E(X) = E(1 * X1 + 2 * X2 + ... + 6 * X6) =
1 * E(X1) + 2 * E(X2) + ... + 6 * E(X6) =
1 * (n/6) + 2 * (n/6) + ... + 6 * (n/6) =
(n/6) * (1 + 2 + ... + 6) =
(n/6) * (6*7/2) = 3.5n

But it doesn't use indicator random variables...