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Two random variables, X and Y, havejoint probability density functionT<y<c+l; 0<a<l f (,y) = {6, otherwise1. Find cvalue. (5 pt) 2. What's the cond...

Question

Two random variables, X and Y, havejoint probability density functionT<y<c+l; 0<a<l f (,y) = {6, otherwise1. Find cvalue. (5 pt) 2. What's the conditional p.d.f of Y given X = 3 , i.e;, fyiX-t (y) ? Don't forget the support of Y. (10 pt) 3. Find the conditional expectation E [YIx]: (10 pt) 4.Find thejoint p.d.f of Xand W = Y _ X.(10 pt) 5.Are V and W are independent? Justify your answer: (5 pt) 6.Find the cumulative distribution function of W. Which named distribution is th

Two random variables, X and Y, havejoint probability density function T<y<c+l; 0<a<l f (,y) = {6, otherwise 1. Find cvalue. (5 pt) 2. What's the conditional p.d.f of Y given X = 3 , i.e;, fyiX-t (y) ? Don't forget the support of Y. (10 pt) 3. Find the conditional expectation E [YIx]: (10 pt) 4.Find thejoint p.d.f of Xand W = Y _ X.(10 pt) 5.Are V and W are independent? Justify your answer: (5 pt) 6.Find the cumulative distribution function of W. Which named distribution is this? (10 pt)



Answers

The joint density function of the random variables $X$ and $Y$ is $$ f(x, y)=\left\{\begin{array}{ll} 6 x, & 0<x<1,0<y<1-x, \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Show that $X$ and $Y$ are not independent. (b) Find $P(X>0.3 \mid Y=0.5)$.

With this expected value, I don't think you need a calculator but calculate would certainly be useful And I just like to think of the outcome with the probability, so the probability of getting a negative five not 0.5 mess up is 0.2. So you have a 20% chance of getting negative five -1 as a 30% chance of .3. A score of zero gives me a point to chance Um and two has a .1 chance And it was next five has a point to chance and I have no idea why there's this number over here because it says that there's no chance of getting this Now your probabilities should add up to be 1.23.5 6 7.8.9 .1 0. So yes it does work. So what you need to do is multiply your probabilities and then add them together. So um you don't have to actually show this math because zero times anything is zero but you are technically multiplying anything and I would say the same thing about that last one that Know anything Times 00. So that's improbable. It's impossible actually. So I'm just double checking my math that makes five times two is negative one, This would be -3. This will be a point to And then five transpoint 1.2 would be one as well. So as you simplify this, I don't know how you would simplify but you don't need a calculator. Um Those negative one and positive one to cancel negative 10.3 plus 0.2 would give me an expected value of negative 0.1 Yeah.


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