My Math Forum joint probability distribution.

 October 27th, 2018, 02:37 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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...
 October 27th, 2018, 07:19 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,404 Thanks: 1306 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 $E[Y]=0.58$ Thanks from zollen
 October 27th, 2018, 07:29 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,404 Thanks: 1306 For (4) you need to include the values. 5) you've done incorrectly. You should end up with 3 element distributions.
October 28th, 2018, 03:08 AM   #4
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Quote:
 Originally Posted by romsek 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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 October 28th, 2018, 07:03 AM #5 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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?
October 28th, 2018, 09:04 AM   #6
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 Originally Posted by zollen 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?

 October 28th, 2018, 09:54 AM #7 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 If you were talking about question 5. No. Let me think about it...
October 28th, 2018, 10:26 AM   #8
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 Originally Posted by zollen 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?

 October 28th, 2018, 11:21 AM #9 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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 }
 October 28th, 2018, 11:43 AM #10 Senior Member     Joined: Sep 2015 From: USA Posts: 2,404 Thanks: 1306 Thanks from zollen

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