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 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: 2,531 Thanks: 1390 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: 2,531 Thanks: 1390 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: 2,531 Thanks: 1390 Thanks from zollen Tags distribution, joint, probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zollen Linear Algebra 0 September 29th, 2018 12:53 PM fun Probability and Statistics 6 January 25th, 2018 04:54 PM IndependentThinker Advanced Statistics 1 February 27th, 2015 02:01 PM xdeathcorex Probability and Statistics 0 April 22nd, 2011 07:55 AM meph1st0pheles Advanced Statistics 1 March 23rd, 2010 05:47 PM

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