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October 27th, 2018, 02:37 PM   #1
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joint probability distribution.

Let X be a random variable with values 0, 1 and 3, and let Y be a random variable with values -1, 1, 2, with the following joint distribution:


Code:
                 Y
           -1     1     2
     ----------------------
     0 | 0.12  0.08  0.10
X   1 | 0.20  0.04  0.25
     3 | 0.08  0.10  0.03



1) Compute the marginal distributions.
	Marg-Dist(X) = { 0.3, 0.49, 0.21 }
	Marg-Dist(Y) = { 0.4, 0.22, 0.38 }

2) Are X and Y independent? Justify your answer.
	No. X and Y are not independent. 
	Recall If X and Y were independent, then P(X,Y) = P(X) * P(Y)
	P(X=1) = P(X=1,Y=-1)+P(X=1,Y=1)+P(X=1,Y=2) = 0.20+0.04+0.25 = 0.49
	P(Y=1) = P(Y=1,X=0)+P(Y=1,X=1)+P(Y=1,X=3) = 0.08+0.04+0.10 = 0.22
	P(X=1,Y=1) = 0.04
	P(X=1) * P(Y=1) = 0.49 * 0.22 = 0.1078
	P(X=1,Y=1) <> P(X=1) * P(Y=1)

3) Compute Exp(X) and Exp(Y)
	Exp(X) = Exp(X=0) + Exp(X=1) + Exp(X=3)
		   = (-1)(0.12) + (1)(0.08) + (2)(0.10) + 
		   	 (-1)(0.20) + (1)(0.04) + (2)(0.25) +
		   	 (-1)(0.08) + (1)(0.10) + (2)(0.03)
		   = 0.58

	Exp(Y) = Exp(Y=-1) + Exp(Y=1) + Exp(Y=2)
		   = (0)(0.12) + (1)(0.20) + (3)(0.08) + 
		   	 (0)(0.08) + (1)(0.04) + (3)(0.10) +
		   	 (0)(0.10) + (1)(0.25) + (3)(0.03)
		   = 1.12

4) Compute the distribution of X + Y.
	= { 0.12, 0.08, 0.10, 0.20, 0.04, 0.25, 0.08, 0.10, 0.03 }

5) Compute the joint distribution of X, Y.
	See Ans 4.

5) Compute P(X|Y=2) and P(Y|X=1).

	P(X|Y=2) = 0.1+0.25+0.03 = 0.38
	P(Y|X=1) = 0.2+0.04+0.25 = 0.49
Would anyone check my math please?? I am not sure I did it correctly...
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October 27th, 2018, 07:19 PM   #2
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You seem to have $E[X]$ and $E[Y]$ flipped.

let $X=(0,1,3)$

and the marginal of $X$ was computed correctly as $p_X=(0.3, 0.49, 0.21 )$

$E[X] = X.p_X = 1.12$

similarly $E[Y]=0.58$
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October 27th, 2018, 07:29 PM   #3
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For (4) you need to include the values.

5) you've done incorrectly. You should end up with 3 element distributions.
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October 28th, 2018, 03:08 AM   #4
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Quote:
Originally Posted by romsek View Post
For (4) you need to include the values.

5) you've done incorrectly. You should end up with 3 element distributions.
Would you provide the correct answers so I know what I did wrong?

Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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October 28th, 2018, 07:03 AM   #5
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Code:
4) Compute the distribution of X + Y.
Answer:
Dist(X + Y) = { P(X=0,Y=-1), P(X=0,Y=1), P(X=0,Y=2),
                     P(X=1,Y=-1), P(X=1,Y=1), P(X=1,Y=2),
                     P(X=3,Y=-1), P(X=3,Y=-1), P(X=3,Y=2) }
              

5) Compute P(X|Y=2) and P(Y|X=1).
Answer:

P(X|Y=2) = { 0.1, 0.25, 0.03 }
P(Y|X=1) = { 0.2, 0.04, 0.25 }
Better?
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October 28th, 2018, 09:04 AM   #6
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Quote:
Originally Posted by zollen View Post
Code:
4) Compute the distribution of X + Y.
Answer:
Dist(X + Y) = { P(X=0,Y=-1), P(X=0,Y=1), P(X=0,Y=2),
                     P(X=1,Y=-1), P(X=1,Y=1), P(X=1,Y=2),
                     P(X=3,Y=-1), P(X=3,Y=-1), P(X=3,Y=2) }
              

5) Compute P(X|Y=2) and P(Y|X=1).
Answer:

P(X|Y=2) = { 0.1, 0.25, 0.03 }
P(Y|X=1) = { 0.2, 0.04, 0.25 }
Better?
does either of those "distributions" sum to 1?
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October 28th, 2018, 09:54 AM   #7
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If you were talking about question 5. No.
Let me think about it...
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October 28th, 2018, 10:26 AM   #8
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Quote:
Originally Posted by zollen View Post
If you were talking about question 5. No.
Let me think about it...
hint - if you have a group of numbers you'd like to cast as a probability distribution how do you usually normalize them?
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October 28th, 2018, 11:21 AM   #9
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0.1+0.25+0.03=0.38
p(x|y=2) = { 0.1/0.38, 0.25/0.38, 0.03/0.38 }

0.2+0.04+0.25=0.49
p(y|x=1) = { 0.2/0.49, 0.04/0.49, 0.25/0.49 }
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October 28th, 2018, 11:43 AM   #10
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