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Two continuous random variables Yi and Yz have the fcllowing joint probability density function312 1 < y2 < V1 < 0; elsewhere_f(y1,y2)(a) Find P(Yi + Yz &l...

Question

Two continuous random variables Yi and Yz have the fcllowing joint probability density function312 1 < y2 < V1 < 0; elsewhere_f(y1,y2)(a) Find P(Yi + Yz < 2)(b) Find the pdf of U =YR.

Two continuous random variables Yi and Yz have the fcllowing joint probability density function 312 1 < y2 < V1 < 0; elsewhere_ f(y1,y2) (a) Find P(Yi + Yz < 2) (b) Find the pdf of U =YR.



Answers

Let $X, Y,$ and $Z$ have the joint probability density function $f(x, y, z)=\left\{\begin{array}{ll}k x y^{2} z, & 0<x, y<1,0<z<2 \\ 0, & \text { elsewhere }\end{array}\right.$ (a) Find $k$. (b) Find $P\left(X<\frac{1}{4}, Y>\frac{1}{2}, 1<Z<2\right)$.

All right. So we've got this joint density function here and were we need to find the value sea. So we know that Ah, this entire density function over the region, eh? Of, ah f of x y z dee's equal to one Because it's a joint density function for end of variables. So we know that we know that one is equal Interval from zero to two to go on reserve two into going on a journey to of C x y z Deac steal I d z. All right, so that sea of the integral and ah, we can just do this as all these separate into girls because they're multiplying and one does not depend on the other. Okay, easy. Okay. And if we just look at one of these into girls X squared over to charity too, we see that that's able to to so that one is equal to C times two to the three. Two c is equal to one of the great. All right, so now we want to look at part be here, so part B, we're looking at everything less than or equal to one settle. We're into girl. Our shuttle into girls are all going to go from one toe zero one that way wth x y z d x y z And very similarly like our previous problem. For the previous part of this problem, at least we can waken you can do the same thing basically right One a into girls You're the one of X d. Jax into girls here to one of why do I do girl zero one of Z D And just looking at one separate one of these, the at one over to so that one over eight times one over two. Cute. It's one over twenty stuck two to the six she and over eight time whenever sixty four. Fantastic. Percy, that explains why Lizzie Lesson or equal to one. You won't find this probability. So we know Z is equal to one minus x minus y. And, uh, what else do we know here? Way. Know that Z is equal to zero. Then Y is equal to one minus X and ah, these are all bounded below zero. Next is around it or above, uh, extended by one. So we're doing this triple in general now, Okay? And one eighth x y Z First we do easy do I and DX eso This is a pretty standard straightforward into girl. So let's go ahead and do this out and ah x y times one minus x minus y squared. Do you know why Dia and ah, I'm actually in line to just leave this the rest of this to you? There's a lot of algebra here in integration, but it's pretty straightforward. It it's not that difficult. Okay, I will write a few of these. I guess you got zero one of one over twelve x to the fifth, minus one over three x to the fourth this one over to certain whenever two x to the third guy won over sixteen. Out here on there's one third X squared minus one twelve packs DX they And once you do this integration, you'LL get one over five seven six zero Ah, as are finally ends

This problem. We've been given the following joint distribution for the variables X, Y and Z. Our first task here is to find the marginal distribution of X and Y. Okay, now we're going to use the fact that these can be shown to be independent in order to break this up and so notice that we can write this as four X. Y times 19 Z swing. Yeah. Now there's a reason I broke the four nights up in that manner. And so that's so if we integrated Z from 0 to 3, that would give us a value of one. And similarly, if we integrate four X. Y over this entire square where both X and y are between zero and one, then a role for X. Y would also give us one. I have been broken apart like this. This tells us that are marginal distribution of action. Y. Mhm. Is able to four X. Y. We're both X and Y. Or between zero and what Now someone really on me where we want the marginal distribution of Z. That's just this other piece out here. This is 1/9 Z squared. Or Z is between zero and three. This is our marginal distribution breezy now and see we want to find the probability of X being between 1/4 and one half. Yeah. Yeah. Uh huh. Okay. Yeah. Of why being greater than one third problem and Z being between two and through. And so we need a triple integral here with extra point from 1/4 to 1 half. Well I going from one third to one because it has an upper bound at one and Z going from 2 to 3. Mhm. Mhm. I'm going to do this is three different intervals because we can break this apart. We can pull the four nines out front and then have the integral from 1/4 to 1 half of X. Dx. Uh huh. We have the integral of one third to one of why do why in the interval from 2 to 3 of Z squared dizzy again just breaking this entire integral part as we have the four nights out front and then we're gonna integrate X from 1/4 to 1 half. And then I will give us a value of 3/32. We're going to integrate why? From one third to one giving us a value of four nights. And then we integrate Z squared from 2 to 3 which is the value of 19/3. And so then we multiply all these together and we do this gives us the value of 19 over 162. Okay. And lastly on C we want a condition R. And D. We want a conditional probability what is the probability the Z. is between zero and two given that both X and Y. Mhm. Are equal to one half. Mhm. Now I'm finding this probability this will be the joint distribution at one half. One half. And then Z going from 0 to 2. All over the marginal of X. Y evaluated at one half. Yeah. Now we get in with the top there, the joint distribution of F. Of one half, one half and Z going from 0 to 2 it means plug in one half for both X and Y. I'm leaving us with 19 Z square and then we'll integrate this from 0 to 2 because we want Z to be between zero and two evaluating the central. It was 8/27. Okay. For the bottom we need F of X Y evaluated at one half one half. So it's a joint distribution that we found right here. We take four times one half times one half, diminish the value of one, so that's one. And so this means that we needed mhm. 8/27. All over one. Which is it paid over 27?

Yeah that's probably been given a falling joint distribution function. I would like to begin by finding the marginal distributions triple X and Y. And the marginal effects just means to integrate out the why? As we integrate from 0-1, three X -Y. over 11. The why? And so this becomes 1/11 times three X y -1/2 White Square. His wife goes from 0 to 1. We put in one for why this is 1/11 Times three X -1/2. So this is our marginal distribution for X. And then for why are marginal for why? This means integrate out the Y about the ex. And so extras from 1-3. So we integrate from 1-3. The three X minus Y over 11. The yes. So this is 1/11 times 3/2 x. where minus Y X. Evaluated from X. is 123. So this is 1/11 times 3/2 times three square minus three Y -1/11 Times three. Have sometimes once where -Y. Times one. And when we combine like terms and simplify this tells us that are marginal distribution of why there's no you have to Y plus 12 all over 11. So that's our marginal for a while now would be we would like to know if X and y are independent and B is a simple no. And that's because we can clearly see that fxfx doesn't I'm sorry, that expects mhm times F. Y of Y does not equal our joint F of X. Y. So for that reason we know they're not independent on C. We want to find the probability that X is greater than two. Now here, all we need to focus on is the marginal for acts And so it's 1 11th Times three X -1/2. And so we just need to integrate this From 2 to 3. Remember X has an upper limit on 3? We don't need to look at the joint, we just need to look at the X. And so we integrated here. He was this 1 11th Times 3/2 expired minus one half X evaluated for Mexico's too just three. So now we're just going to plug in three Party into and then subtract. I don't know if we do. This gives us a value of seven over 11.

Yeah. It's probably want to find the marginal distributions given the joint probability distribution here. Now in order to find a marginal distributions, we need to integrate out the other herb. So F X with X. It means we integrate out the Y. That's we integrate from 0 to 1 three x minus Y over 11. Do you want? And so this gives us 1/11 times three X y minus one half. Y squared evaluated from y is zero to one point in 1.10 for Y. And then some track this gives us 1/11 times three x minus one half is our alphabet. Okay. Now, similarly for F some Y of Y we integrate out correct. And so this is 1/11 times three halves that squared minus X. Y. Evaluated from articles 123 Yeah. And so now for actually going to plug in one, we're going to plug in three and then we're going to subtract them. And whenever we do this gives us 1/11 times 12 minus two. Y besides our marginal distribution. For why? Mhm. No. And be we want to discuss that these are independent but clearly we can see that our joint distribution is not equal to the product of our marginal distributions. In order for them to be independent, the joint distribution must be equal to the product of the marginal distribution. And so since this is not true, they are not independent. No. And see we want to find the probability that X is greater than two. And so this is just the integral from 2 to 3 of just our X distribution. We do not need the Y. We just need the X distribution. And so we'll integrate from 2 to 3. This marginal that we just found, which is 1/11 times three X minus a half. So our anti derivative here, it will be 1/11 times three has X squared minus one half X evaluated from access to 23 And so then now we're going to plug into 40 in three and then some traveling. And or if we evaluate this sonia was seven overall up. Yeah.


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