How do I derive the poisson distribution?

2. Originally Posted by bluebiro
Say D is distributed by N(2,2), E is distributd by N(3,3) and F is distributed by N(4,4), and D, E and F are independent.

(a) Find E(D squared plus E squared plus F squared)
I get 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29
By linearity of the expectation operator:

$E(D^2+E^2+F^2)=E(D^2)+E(E^2)+E(F^2)$

Now:

$var(X)=E(X^2)-\overline{X}^2$

so:

$E(X^2)=var(X)+\overline{X}^2$

CB

3. Originally Posted by bluebiro
Say D is distributed by N(2,2), E is distributd by N(3,3) and F is distributed by N(4,4), and D, E and F are independent.
(b) Find Cov(D+E, D-2E)
(stuck)
$Cov(D+E, D-2E)=E( [(D+E)-(\overline{D}+\overline{E})][(D-2E)-(\overline{D}-2\overline{E})]$

Now expand the expression inside the expectation on the right hand side. Then use linearity of the expectation operator to express the right hand side as a sum of expectations, and use the fact that the means and variances of all the RV's are known and that they are independent to expand the resulting expression.

CB

4. Originally Posted by bluebiro
Say D is distributed by N(2,2), E is distributd by N(3,3) and F is distributed by N(4,4), and D, E and F are independent.

(c) What is the resulting Normal distribution of D+E+F?
I get N(0, 6)
How do you get that, if you explain we can see what the problem is with your understanding and so help you.

CB

5. (b) Cov(D+E, D-2E)=Cov(D,D)+Cov(E,D)-2Cov(D,E)-2Cov(E,E)=V(D)+0+0-2V(E)

6. Originally Posted by CaptainBlack
How do you get that, if you explain we can see what the problem is with your understanding and so help you.

CB
Sorry, I don't know what I was doing there.

I get (2,12) as the new distribution this time?

7. Originally Posted by bluebiro
Sorry, I don't know what I was doing there.

I get (2,12) as the new distribution this time?
My question is: What are you doing to get this answer (if you are guessing tell us you have no idea what to do)?

CB

8. Originally Posted by bluebiro
Sorry, I don't know what I was doing there.

I get (2,12) as the new distribution this time?
Read this: Sum of normally distributed random variables - Wikipedia, the free encyclopedia

9. Originally Posted by CaptainBlack
My question is: What are you doing to get this answer (if you are guessing tell us you have no idea what to do)?

CB
Sorry, my original question was typed wrong. I wanted to find the resulting distribution of D+E-F...

10. Originally Posted by bluebiro
Sorry, my original question was typed wrong. I wanted to find the resulting distribution of D+E-F...
and my working was:

E(new function)=3+4-5=2
Var(new function)=3+4+5=12

11. Originally Posted by bluebiro
and my working was:

E(new function)=3+4-5=2
Var(new function)=3+4+5=12
Which looks OK.

CB

12. Originally Posted by matheagle
(b) Cov(D+E, D-2E)=Cov(D,D)+Cov(E,D)-2Cov(D,E)-2Cov(E,E)=V(D)+0+0-2V(E)
Why is the covariance (E,D) = 0 ?

13. Originally Posted by bluebiro
Why is the covariance (E,D) = 0 ?
$E$ and $D$ are independent RV's and:

$Covar(E,D)=E((E-E(E))(D-E(D)))=E(E-E(E)) E(D-E(D))$

and $(E-E(E))$ and $(D-E(D))$ are zero mean RV's.

CB

14. Originally Posted by CaptainBlack
$E$ and $D$ are independent RV's and:

$Covar(E,D)=E((E-E(E))(D-E(D)))=E(E-E(E)) E(D-E(D))$

and $(E-E(E))$ and $(D-E(D))$ are zero mean RV's.

CB
(sorry for all the questions)

What does "zero mean RV's" mean?

15. Originally Posted by bluebiro
(sorry for all the questions)

What does "zero mean RV's" mean?
RV :- Random Variable

"zero mean":- has a mean of zero.

CB

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