Probability, Correlation, Variance Statistics Homework Help.

[JasonBrewer]

Homework Statement

1. Assume each birth in a hospital on a given day is independent of one another, and each birth, P(boy)=0.48. What is the probability that the 8th baby born is the 5th girl.

2. Two random variables X and Y have joint distribution given by. What is their correlation.

3. Z is drawn randomly from {1,2,3,4,5}. Independently of Z, 4 other RVs Y1,...,Y4 are drawn randomly, with replacement, from {1,...,6}. Define X as the number of Yjs less than or equal to Z. What is the variance of X?

Homework Equations

2. Corr(X,Y)=Cov(X,Y)/((Var(X),Var(Y)^0.5)
where Cov(X,Y)=E(X,Y)-E(X)E(Y)

The Attempt at a Solution

I think that most of my problem comes from not being able to apply the formulas and theory since my teacher doesn't do very many problems, and mainly does proofs which I don't find too helpful.

Anyways my theory.
1. So I think that I need to find the probability that out of the first 7 babies, 4 are girls, and I'm almost certain that this is a binomial. But from there, I don't know what to do. Then multiply by the chance that the 8th born, is a girl (0.52).
Edit: I got this one! I found the forumla for probability and applied it.

2. This is an issue what I don't know how to intepret the data. To find Var, I need to Expectation, but how do I find it? Which numbers do I multiple to get it.
I also need to find Cov(X,Y) which require E(X,Y), which I am also clueless to find when the data is in this form.
Is it E(X)=-1*(0.1+0.1+0.1)+0(0.1+0.15+0.1)+1(0.15+0.1+0.1)
and E(Y)=-1(0.35)+1(0.3)
When I tried this, the answer was almost certainly incorrect, since I got a positive answer, and the real answer I know is -ve.
I thought that maybe the above might be correct, and I just don't know how to compute E(X,Y) correctly.

3. This last one, I was given a hint. Var(X)=var[4z/5]+E((4z/5)(1-(z/5))
I still don't know how to do this question even with this forumla. I assume that there are only a couple more steps (2?), but I don't know how to continue with the question.

Thanks for any help

 

[haruspex]

Is it E(X)=-1*(0.1+0.1+0.1)+0(0.1+0.15+0.1)+1(0.15+0.1+0.1)
and E(Y)=-1(0.35)+1(0.3)

Yes

When I tried this, the answer was almost certainly incorrect, since I got a positive answer, and the real answer I know is -ve.

So post your working.

I thought that maybe the above might be correct, and I just don't know how to compute E(X,Y) correctly.

You mean E(XY). It's the same way you calculate the expected value of any r.v. List the values of XY, multiply each by its probability, and add them up.

3. This last one, I was given a hint. Var(X)=var[4z/5]+E((4z/5)(1-(z/5))

That's rather a specific hint. I mean, it's unclear how to adapt that to a different question. Anyway, not sure what's stopping you from applying it.
What's the relationship between var(aZ) and var(Z) for a constant a?
How can you expand E(aZ(b-cZ))?
(Btw, you should be using capital Z inside Var() and E(), as I have, because these refer to random variables. The lower case letters are used for individual values of the random variables.)

 

[JasonBrewer]

YesSo post your working.You mean E(XY). It's the same way you calculate the expected value of any r.v. List the values of XY, multiply each by its probability, and add them up.
That's rather a specific hint. I mean, it's unclear how to adapt that to a different question. Anyway, not sure what's stopping you from applying it.
What's the relationship between var(aZ) and var(Z) for a constant a?
How can you expand E(aZ(b-cZ))?
(Btw, you should be using capital Z inside Var() and E(), as I have, because these refer to random variables. The lower case letters are used for individual values of the random variables.)

Sorry you were correct for E(XY)
With E(XY), how would you do the two together?
Would it be something like:
(-1*-1*0.1)+(-1*0*0.1)+(-1*1*0.1)+...+(1*1*0.1)?

With that hint, it's meant specifically for this questions. Anyhow.
Do you mean for a constant a, it can be brought to the find of it?
E(aZ(b-cZ))=aE(Z(b-cZ))?

Thanks for replying, I'm really struggling with the subject.

 

[haruspex]

With E(XY), how would you do the two together?
Would it be something like:
(-1*-1*0.1)+(-1*0*0.1)+(-1*1*0.1)+...+(1*1*0.1)?

Yes

With that hint, it's meant specifically for this questions. Anyhow.
Do you mean for a constant a, it can be brought to the find of it?
E(aZ(b-cZ))=aE(Z(b-cZ))?

Yes. E is a linear function. E(aX+bY) = aE(X)+bE(Y).
But be careful with Var - that is not linear. To see what happens to Var(aX), use the formula Var(X) = E(X²)-E(X)².

 

[JasonBrewer]

Yes
Yes. E is a linear function. E(aX+bY) = aE(X)+bE(Y).
But be careful with Var - that is not linear. To see what happens to Var(aX), use the formula Var(X) = E(X²)-E(X)².

Ok so i know I'm meant to use the forumla Corr(X,Y)=E(XY)-E(X)E(Y) / sqrt[Var(X)Var(Y)]

What I've got so far
E(X)=0.05
E(X^2)=0.65
E(Y)=-0.05
E(Y^2)=-0.65
Var(X)=Var(Y)=0.6475
E(XY)=-0.05
When I put it into the forumla, the answer the I get isn't correct.
Is there something that I'm doing wrong?

For the last question, I'm still not sure what I'm meant to do. What am I looking for?
E(Z)=(1*0.2)+(2*0.2)+3*(0.2)+(4*0.2)+(5*0.2)??
What am I meant to apply to this Z/5?

Thanks

 

[haruspex]

Ok so i know I'm meant to use the forumla Corr(X,Y)=E(XY)-E(X)E(Y) / sqrt[Var(X)Var(Y)]

I trust you mean (E(XY)-E(X)E(Y)) / sqrt[Var(X)Var(Y)]

What I've got so far
E(X)=0.05
E(X^2)=0.65
E(Y)=-0.05
E(Y^2)=-0.65
Var(X)=Var(Y)=0.6475
E(XY)=-0.05
When I put it into the forumla, the answer the I get isn't correct.
Is there something that I'm doing wrong?

Looks correct to there, so you'll have to post the rest of your working. I get -0.073.

For the last question, I'm still not sure what I'm meant to do. What am I looking for?
E(Z)=(1*0.2)+(2*0.2)+3*(0.2)+(4*0.2)+(5*0.2)??

Yes

What am I meant to apply to this Z/5?

Not yet. Expand E((4Z/5)(1-(Z/5)) first, using the linearity formula I gave you.

 

[JasonBrewer]

I trust you mean (E(XY)-E(X)E(Y)) / sqrt[Var(X)Var(Y)]

Looks correct to there, so you'll have to post the rest of your working. I get -0.073.
Yes
Not yet. Expand E((4Z/5)(1-(Z/5)) first, using the linearity formula I gave you.

OH MY LORD. My friend gave me an answer to the wrong question! -0.073 was correct! I thought that I waas wrong the whole time. Thanks a bunch :D

Var(X)=Var(4Z/5)+4E[z(1-(z/5))]
For Var(4Z/5), you said that it doesn't scale linearly, and to use E(X) and E(X^2), but how do I apply either of them in this situation?

Thanks

 

[haruspex]

Var(X)=Var(4Z/5)+4E[z(1-(z/5))]
For Var(4Z/5), you said that it doesn't scale linearly, and to use E(X) and E(X^2), but how do I apply either of them in this situation?

I need to some attempt from you here.
Var(X) = E(X²)-E(X)².
Suppose Y=aX, some constant a.
So what do you get for Var(Y)?

 

[JasonBrewer]

Ok so I'm not sure if this is correct, but I'll try and explain my logic behind this.
So X is the number of Ys Less than or equal to Z.
For Z=1, there is only one thing thess than or =to 1 (1/6) from Y, which is one. For 2, there are 2 things, 1 and also 2.
so I get
E(X)=(1*1/6)+(2*2/6)+(3*3/6)+(4*4/6)+(5*4/6)+(6*4/6)
E(X^2)=(1*1/6)+(4*2/6)...+(25*4/6)+(36*4/6)

 

[JasonBrewer]

For 4E(Z/5 - Z^2/25)
E(Z/5-Z^2/25)=(1/5 * 1/5)+(2/5 * 1/5)+...+(5/5 * 1/5)+ [(1^2/25) *1/5]+[(2^2/25)*1/5] +[(3^2/25)*1/5]+...+[(6^2/25)*1/5]
Then multiply that above answer by 4.
And this ans my post above, that would be the answer?

 

[haruspex]

Ok so I'm not sure if this is correct, but I'll try and explain my logic behind this.
So X is the number of Ys Less than or equal to Z.
For Z=1, there is only one thing thess than or =to 1 (1/6) from Y, which is one. For 2, there are 2 things, 1 and also 2.
so I get
E(X)=(1*1/6)+(2*2/6)+(3*3/6)+(4*4/6)+(5*4/6)+(6*4/6)

No, it's rather more complex than that.
E.g. suppose Z=3. What is the probability that none of the Y are <= 3? That is the probability P(X=0|Z=3). It would be a lot of work to write out all 25 combinations of P(X=x|Z=z), so see if you can work out the general formula.

 

[haruspex]

For 4E(Z/5 - Z^2/25)
E(Z/5-Z^2/25)=(1/5 * 1/5)+(2/5 * 1/5)+...+(5/5 * 1/5)+ [(1^2/25) *1/5]+[(2^2/25)*1/5] +[(3^2/25)*1/5]+...+[(6^2/25)*1/5]
Then multiply that above answer by 4.

Yes, except that you turned a minus into a plus. Don't forget the Var[4z/5] term of the hint.
Btw, I don't know how that hint formula was derived, and it gives me a different answer than the one I get if I don't use it. But I may have made a mistake, of course.

 

[JasonBrewer]

Is the formula that I'm meant to be using nCr*p^r*(1-p)^(n-r)?

 

[JasonBrewer]

Yes, except that you turned a minus into a plus. Don't forget the Var[4z/5] term of the hint.
Btw, I don't know how that hint formula was derived, and it gives me a different answer than the one I get if I don't use it. But I may have made a mistake, of course.

I have the answer to the question. Could you tell me the anser that you get. If you get different answers for the hint method and the method that you have chosen to do, I don't want to be doing something incorrectly.

 

[JasonBrewer]

Hint is derived from
X|Z=z - B(4,(Z/5))
Var(X)=Var[E(X|Z)]+E[Var(X|Z)]
so Var(X)=[4Z/5]+E[4(Z/5)(1-(Z/5))]

 

[haruspex]

Is the formula that I'm meant to be using nCr*p^r*(1-p)^(n-r)?

If not using the hint, yes.

 

[JasonBrewer]

Could you tell me if the answer that you got without/without the hint is 1.67?

 

[haruspex]

Hint is derived from
X|Z=z - B(4,(Z/5))

You mean X|(Z=z) = z - B(4,(z/5))?
That doesn't look right to me. E.g. P(X=4|Z=1) = 6^-4. The RHS is not going to give a power of six in the denominator.

Var(X)=Var[E(X|Z)]+E[Var(X|Z)]
so Var(X)=[4Z/5]+E[4(Z/5)(1-(Z/5))]

Using the hint I get 48/25. What do you get? I'm fairly certain the right answer is a bit less.

 

[JasonBrewer]

You mean X|(Z=z) = z - B(4,(z/5))?
That doesn't look right to me. E.g. P(X=4|Z=1) = 6^-4. The RHS is not going to give a power of six in the denominator.

Using the hint I get 48/25. What do you get? I'm fairly certain the right answer is a bit less.

I wasn't sure how to work out a general formula for use with the hint, but if you got 48/25, I doubt that you're wrong, and suspect that the hint is incorrect-which is a bummer.
How would you calculate the answer without the hint?

The reason that I used this "Is the formula that I'm meant to be using nCr*p^r*(1-p)^(n-r)?", was because I thought that it was a binomial since it's with replacement.

 

[JasonBrewer]

Oh if you missed the answer on the previous page, it was 1.67 so you were right in thinking that it was less.

 

[haruspex]

Could you tell me if the answer that you got without/without the hint is 1.67?

Yes. I suspect the hint method should be using z/6, not z/5.

 

[JasonBrewer]

Yes. I suspect the hint method should be using z/6, not z/5.

Would you say that the method that you tried easier than the method given here by the hint?

 

[JasonBrewer]

Also would that mean that:
E(Z/6-Z^2/36)=(1/5 * 1/6)+(2/5 * 1/6)+...+(5/5 * 1/6)+ [(1^2/36) *1/5]+[(2^2/36)*1/5] +[(3^2/36)*1/5]+...+[(6^2/36)*1/5] ?

 

[haruspex]

Also would that mean that:
E(Z/6-Z^2/36)=(1/5 * 1/6)+(2/5 * 1/6)+...+(5/5 * 1/6)+ [(1^2/36) *1/5]+[(2^2/36)*1/5] +[(3^2/36)*1/5]+...+[(6^2/36)*1/5] ?

Yes, except that, as I pointed out in an earlier post, you've changed a minus to a plus.
E(Z/6 - Z^2/36).
I confirm that using z/6 instead of z/5 throughout the hint gives the right answer.

 

[JasonBrewer]

Could you tell me a method that I can use without the hint?
Althought the hint is helpful, I doubt that I will eb able to remember it.
And yes I forgot to change the minus sign, I just copied and pasted it and just changed the denominators.
E(Z/6-Z^2/36)=(1/5 * 1/6)+(2/5 * 1/6)+...+(5/5 * 1/6)-{ [(1^2/36) *1/5]+[(2^2/36)*1/5] +[(3^2/36)*1/5]+...+[(6^2/36)*1/5] }?

I'm unable to figure out a formula for E(X) so I'm looking for an alternate method.

 

[JasonBrewer]

Is the general forula ((Z+1)/6)^4?
E.g. suppose Z=3. What is the probability that none of the Y are <= 3? That is the probability P(X=0|Z=3). It would be a lot of work to write out all 25 combinations of P(X=x|Z=z), so see if you can work out the general formula.
Since it would need to be greater than 3. Then since we're sampling with replacement, the denominator is always 6. Then to the power of 4 for the number of draws.

 

[haruspex]

Is the general forula ((Z+1)/6)^4?

No, it's going to look like ⁿCr*p^r*(1-p)^n-r. You just have to figure out what to substitute for p, n and r.
First, n: how many things are we comparing with z?
Next, p: If Z=z, and a given Y is from 1 to 6, what is the probability that Y <= z?
(Try z=1, etc.)
Now r: what is the probability that r of the Y are <= z? How does r relate to X?

 

[JasonBrewer]

No, it's going to look like ⁿCr*p^r*(1-p)^n-r. You just have to figure out what to substitute for p, n and r.
First, n: how many things are we comparing with z?
Next, p: If Z=z, and a given Y is from 1 to 6, what is the probability that Y <= z?
(Try z=1, etc.)
Now r: what is the probability that r of the Y are <= z? How does r relate to X?

Thanks haruspex for being patient and helping me out with the 2 questions. You've been a real help and I can't thank you enough for how helpful you have been.
I hope you're around when I have more statistics related questions in the near future.