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S6.5 Let (XY) be continuous random vector with joint probability density function1/2 fxye if -1 < 1 < [ and 0 < y < 1 ~(x.y) = { otherwiseThen P(X > ...

Question

S6.5 Let (XY) be continuous random vector with joint probability density function1/2 fxye if -1 < 1 < [ and 0 < y < 1 ~(x.y) = { otherwiseThen P(X > 0) equals (A) 3*12 D4

S6.5 Let (XY) be continuous random vector with joint probability density function 1/2 fxye if -1 < 1 < [ and 0 < y < 1 ~(x.y) = { otherwise Then P(X > 0) equals (A) 3*12 D4



Answers

The joint density function for random variables $ X $, $ Y $, and $ Z $ is $ f(x, y, z) = Cxyz $ if $ 0 \le x \le 2, 0 \le y \le 2, 0 \le z \le 2 $, and $ f(x, y, z) = 0 $ otherwise.
(a) Find the value of the constant $ C $.
(b) Find $ P(X \le 1, Y \le 1, Z \le 1) $.
(c) Find $ P(X + Y + Z \le 1) $.

Well that's probably been given the following joint probability distribution. And I would like to first show that they are not independent now in order for them to be independent. So that's why must be equal to the product of the marginal distributions of the two. So in order to do this let's find the marginal distributions of the tune and from that are marginal distribution effects means we integrate out the why? So why does from 0 to 1 -X. So this is three X squared evaluated from zero. I'm sorry not three X squared be integrated with respect to why not with respect to X integrate with respect to why? This is to be six x. Y Evaluated from X0 To one -X. So this is six x Times one -X. This is our marginal distribution of X. Yeah. The marginal distribution of why we integrate out the X and two X goes from 01 Now we will integrate with respect to Act this is three X squared yeah Evaluated from 0 to 1. And so this will give us A value of three and so are marginal distribution for why It's just a value of three. Now clearly we can see. Mhm. That f of why why times F. Of X. XR two marginal distributions is not equal to our joint distribution. Thus that's why or not independent and that's what we need to show. Their army. Yeah on me We want to find the probability that that's better than a .3. What given that why is equal 0.5? In order to evaluate this this is equal to the probability That it's just larger than .5. And Why is 0.5 over The probability that why is 0.5 using a conditional probability laws? I'll talk there. That's just our joint distribution. And so our joint distribution is six X. That's where we get this on top and then on the bottom that's our marginal distribution are marginal distribution for why? Those are three. So this gives us this distribution here to be two X. Now we would plug in 0.5 for why? But there is no way to plug it into here. Mm We do want us to be bigger than .5. And so we integrate this From 05 to one because X has an upper limit one. So this is x squared evaluated from 0.5. To wonder That's why party on one With 41.5. And subtract This gives us .75.

Bullet and grow here will always start with the inner integral. And that's from 0 to 1 with respect to X. So we're going to treat why, like a constant And we'll go ahead and just anti derived that right now. So they give us execute divided by three. Plus, I'm gonna go ahead and keep the Constance as red one third, why? And then anti derivative of X, which is X squared over two. So that, of course, just gives us over six. Here is the constant. Okay. And then we want to in a evaluate this rather from you just say over one 0 to 1. And all of that is what is in the green brackets here. So we're just going to come up with one more step and then we still need to integrate that with respect to why, from 0 to 2, Okay, if we plug in zero into this expression here, that would, of course, give us zero. So we really just need to plug in one and remembering that we're substituting for X here. So this will give us one third and then plus this'll just give us one sex. Why? So this is what we are now anti deriving with respect. Took why from 0 to 2. Okay and go ahead and do that in the next step. Here. Let's go ahead and keep it and blue here. So this would give us one third anti derivative of a constant is that times the variable plus 1/6 anti derivative of Y y squared over two. Let's go ahead and change that down there to a 12. Then that just got multiplied. And then we're gonna evaluate this from 0 to 2. Same thing zero would result values that also we're zero. So we really just need to plug to in. So finally, this will give us two thirds Waas Xi squared 4/12 for over 12 is just the same thing as one third two thirds plus one third finally just simplifies you want. And so that's what this double undergrowth evaluates dio

Well that's probably been doing the following joint distribution. Now, the first thing we invite to find is the value of K and find the value of day. We know that the triple integral of K X y squared Z Must be equal to one. In order for us to a probability distribution. No extras from 0 to 1. Why goes from 0 to 1 and z It goes from 0 to 2. This means we have K times the integral from 0-2, brazil is easy. The integral from 0 to 1 of y squared. Dy the admiral From 0 to 1 of X dx Is equal to one. Now the integral From 0 to 2 of Z. Z is equal to The integral from 0 to 1 of why squared is one third And the integral from 0 to 1 Of that is 1/2. Yeah. Mhm. Mhm Two and one half canceled. So this tells us that K over three is one. Yeah, That's OK. Is equal to three. Yeah. Some things are joint distribution is really F of X. Y. Z. Mhm. This three X Y squared Z. With X&Y. Between zero and 1 Lindsey. Between zero and 2. That it helps us on B. Because one of the probability That acts as less than 1/4, Why is greater than 1/2 And Z is between one and 2. So finding this probability we want to integrate the triple integral of three at Weiss Birds. E Z goes from 1 to 2, Hind rose from 1/2 to 1 because it has an upper bound at one. Then Ash goes from 0 to 1/4 because it has a lower bound at zero. So this is equal to three times the integral From 0 to 1 4th versus the arts Times. The integral from 1/2 to 1 of Y squared dy Yeah, I'm gonna go from 1 to 2 of Z dizzy. So this is three times now. If we end great From 01 4th of X. Dx We get 1/32. If we integrate, why squared From 1/2 to 1, What do you 7/24? And then if we integrate Z DZ from one to We get three hands and then we multiply all these together. Mhm. This is 21/5, 12. Yeah.

In this question were given a joint density function And we want to verify that this is a first that this is a probability density function. Second, we want to calculate a few probabilities and third we want to find the expected values of both variables. We're assuming that both of the variables are discreet that they can take on any value. So how do we show that this is a valid pdf or a probability density function? We can do that by. We do that by making sure that the double integral over all of the reels. Yeah. Mhm. We need to make sure that this entire function Has an integral an area under this plane of one. So how do we do that with this question? Well that would mean that the that both double integral from negative infinity to infinity. I would have to be one. Fortunately We only have to go through one side because the function is only non-0 if x and y are both positive. So this means that what we actually have to check, we actually have to check the value of the integral from zero to infinity. Of the whole exponential function. Yeah. So you just have 0.1 times the exponential function of negative 0.5 X minus 0.2. Why? And thankfully in this case it doesn't matter the order in which you put your differentials. The integral can be split according to the according to the multiplication role of exponential. So have 0.1 times east, the -0.5 x D X plus the double integral Of 0.1 times e. to the -0.2. Why? Right? Yeah. And by this point you know how to do already. Mhm. We know how to do these integral is already and we actually only need one integral sign for each one. We know how to do improper integral. So evaluating each integral will give us You'll have 0.1 Over negative 0.5 times east, the -0.5 X. That goes from 0 to infinity. And the same thing. Yeah. Mhm. Mhm. The same thing there. Now the infinity terms when I plug infinity and will just vanish And this will just give us 0.1 times zero point time 0.5. Yes. Mhm. Plus 0.1 times one over. Don't forget these are actually being multiplied. I've made a mistake there. I made a mistake that there should be multiplied together because of the product rule for integral. So And because this is one we know that this is a valid probability density function. Yeah. I'll just write down the theorem that I used here. Okay, now the second part is we want to find two probabilities. You want to find the probability that why Our random variable y is greater than one. So that's the first step. Okay, how do we do this? Well all we need to do is we just need to change our bounds of integration a little bit. Yeah. So this will turn out to be the integral from zero to infinity of 0.1 times the X times E to the minus 0.5 X. Being multiplied by the integral from one to infinity of 0.1 times E. To the minus 0.2. Why it? And by similar logic We have that by similar logic to how it did the integral above. We have that this is going to be equal to approximately 0.8187. Yeah. Now the second part is we want that the probability mhm But our random variable X is less than two And at the same time I random variable why is less than four. That means that we're going to have our integral as going from 0-2 and from 0-4. And I can say that it's just going to set the beginning band will be zero because everywhere else the function is zero. So that will be the integral from 0 to 1. From 0 to 2 of 0.1 times each -0.5 x. Multiplied by the integral from 0 to 40.1 times E to the minus 0.2. Why? And doing these intervals in much the same way as we did before Will result in a probability of 0.3481. Yeah. Yeah. Mhm. Now the last part is we want to find the expected values of X and y. So we want mhm We want those values. So how do we find those? Well, the expected value of each integral. Yeah. Which will note down here, it's just the double integral over the real plain of why? Mhm. Yeah. And we'll have an X be multiplied by it. And the expected value for why similarly is going to be yeah the integral over the real plain of why times f of times the function mm So let's begin finding the expected value of x. 1st we know that this is going to be equal to X Will be the end score from 0 to Infinity and score from 0 to Infinity. And we'll have 0.1 Times X Times East zero just like that. Which we can again split into an integral from multiplication. Yeah. Okay. Yeah. In fact and the first integral we can do by parts and by similar logic. Yeah. Because of the first and then second integral. We can just do that on its own. Uh huh. The first one will require integration by parts but when we do so we're going to get that. This is gonna be too Okay. And similarly I'll just skip to the other to the part where we like to see it. The expected value for why It's going to be just 25 or that will be five Because of the same logic. What? And that's how you do this question


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