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 Probability Old Exam P Forum

#141
07-11-2008, 03:38 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 6,155

Assuming you didn't know about it, though you probably should as it will certainly pop up in 4/C, here's a way to do a problem in an equivalent way.

Assuming you know that V(X) = E(V(X|Y)) + V(E(X|Y)) let's calculate the 2 pieces we are summing

$E(V(X|Y)) \, = \, \frac{1}{3} (1) \, + \, \frac{2}{3} (4) \, = \, 3$

since we are just treating V(X|Y) as a R.V. and finding the expected value by multiplying outcomes by prob. of those outcomes

$V(E(X|Y)) \, = \, E([E(X|Y)]^{2}) \, + \, (E[E(X|Y)])^{2}$

You should know this simply because you should know that V(Blah) = E(Blah^2) - [E(Blah)]^2, well here Blah is E(X|Y).

Now,

$E[E(X|Y)] \, = \, \frac{1}{3} (2) \, + \, \frac{2}{3} (4) \, = \, \frac{10}{3}$

and

$E[E(X|Y)]^2 \, = \, \frac{1}{3} (2)^{2} \, + \, \frac{2}{3} (4)^{2} \, = \, 12$

therefore

$V(E(X|Y)) \, = \, 12 \, - \, \left(\frac{10}{3}\right)^{2} \, = \, \bar{.8}$

and finally

$V(X) \, = \, 3 \, + \, \bar{.8} \, = \, 3.\bar{8}$
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.

Last edited by Actuarialsuck; 07-18-2008 at 10:57 AM..
#142
07-12-2008, 01:05 AM
 gametime50 Member Join Date: May 2008 Posts: 82
May 12/08 question

Could you remind me again what that shortcut is? I have not seen or heard of it i just know the variance of the bernoulli being pq?
#143
07-12-2008, 09:14 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 6,155

Quote:
 Originally Posted by gametime50 Could you remind me again what that shortcut is? I have not seen or heard of it i just know the variance of the bernoulli being pq?
Suppose you have a R.V. X that has outcome x with prob. p(x) and outcome y with prob. p(y).

Then $V(X) \, = \, \left(x \, - \, y \right)^{2} p(x) p(y)$

You can derive this formula just by computing moments and knowing your algebra

Spoiler:

$E(X) \, = \, x p(x) \, + \, y p(y)$

and

$E(X^{2}) \, = \, x^{2} p(x) \, + \, y^{2} p(y)$

therefore since

$V(X) \, = \, E(X^{2}) \, - \, [E(X)]^{2} \, = \, x^{2} p(x) \, + \, y^{2} p(y) - \left[x^{2} {p(x)}^{2} \, + \, 2xy p(x) p(y) \, + \, y^{2} {p(y)}^{2} \right] \, = \,$

$x^{2} p(x) \left[ 1 \, - \ p(x) \right] + y^{2} p(y) \left[1 \, - \, p(y) \right] \, - \, 2xy p(x) p(y) \, = \, x^{2} p(x) p(y) \, + \, y^{2} p(x) p(y) \, - \, 2xy p(x) p(y) \, = \, \left( x \, - \, y \right)^{2} p(x)p(y)$
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.

Last edited by Actuarialsuck; 07-12-2008 at 09:19 PM..
#144
07-14-2008, 04:41 PM
 gametime50 Member Join Date: May 2008 Posts: 82
May 12/08 question

Thank you for the response. I wander how often that will show up for exam P? Im just starting the testing process so I havent seen any of the stuff beyond Exam P really. I think you meant to have a couple - signs in there instead of + but i got what you meant. Thanks again.
#145
07-14-2008, 04:43 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 6,155

Where?
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.
#146
07-15-2008, 12:37 AM
 gametime50 Member Join Date: May 2008 Posts: 82
May 12/08 question

in the blah parts
#147
07-15-2008, 12:41 AM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 6,155

Quote:
 Originally Posted by gametime50 in the blah parts
What do you mean? Point to actual places in the derivation.
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.
#148
07-15-2008, 07:36 PM
 gametime50 Member Join Date: May 2008 Posts: 82
may 12/08 question

No you did the problem fine i think you made a typo when you had the v(blah)= E(blah^2) + instead of - (E(blah))^2
#149
07-18-2008, 09:35 AM
 sam_broverman Member Join Date: Feb 2005 Posts: 639

Actuarialscuk,

Thanks for responding to the question. I was
out of internet contact for a little while.

Sam Broverman
#150
07-18-2008, 10:57 AM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 6,155

Quote:
 Originally Posted by gametime50 No you did the problem fine i think you made a typo when you had the v(blah)= E(blah^2) + instead of - (E(blah))^2
I finally saw what you were talking when I realized I actually used the word "blah". Thanks, I corrected it.
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.