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(# 5) Let Y & Z be independent random variables each of which are uniformly distributed over (-2,2). Find the probability that the equation x? + 0SYx + Z=0 has ...

Question

(# 5) Let Y & Z be independent random variables each of which are uniformly distributed over (-2,2). Find the probability that the equation x? + 0SYx + Z=0 has real roots (in x)

(# 5) Let Y & Z be independent random variables each of which are uniformly distributed over (-2,2). Find the probability that the equation x? + 0SYx + Z=0 has real roots (in x)



Answers

Suppose that $X$ and $Y$ are independent random variables with probability densities and $$ g(x)=\left\{\begin{array}{ll} \frac{24}{x^{4}}, & x>2 \\ 0, & \text { elsewhere } \end{array}\right. $$ and $$ h(y)=\left\{\begin{array}{ll} 2 y, & 0<y<1 \\ 0, & \text { elsewhere } \end{array}\right. $$ Find the expected value of $Z=X Y$.

Yeah, that's probably been doing this distribution function and we all like to find the expected value of route X squared false watch where the expected value of root X squared plus Y squared is going to be the double integral of route X squared plus Weisberg times or probability distributions of times for X. Y. The arts do I. Next is greater than zero. And so extras from zero to infinity. Yeah. Why is less than one and two? Why does from negative infinity? Mhm. Up to one. That's what we're wanting to evaluate this interval. Because our bounds are not dependent upon any of the variables. We can rewrite it. So this will be the integral from the of insanity to one four. While we can bring those out times in a row from zero to infinity of eggs. Route next word. Both wash word. Yeah. Taking our anti derivative here will still have this negative infinity to one on the outside this before. Why times? Okay, X squared plus Y squared to the three house over three evaluated for magical zero, two and three. A little misunderstanding my apartment said zero is less than X. Comma, Y is less than one. That means they're both between your own one. So I need to change these bounds here and you go from 0 to 1. Sorry about that 0 to 1. And both of these. I mean they're both between your own one. That'll make this work out better for us here. So extras from 0 to 1. So now we're going to evaluate that we put in one. We put in zero and then we subtract them. And in doing this, yeah. We're left with the integral from 0 to 1 of four Y. Time one of the third over three. So that's what we get when we plug in zero in the fucking one. Yeah. We have Y squared plus one 23 house over three minus. Why do the third over three dy distributed in the four Y. Means we have the integral from 0 to 1 of four thirds. Y. Times widespread plus one to the three house minus four thirds. Why do the fourth dy take your anti derivative now? Mhm. And this will give us 4 15 widespread plus one. The five halves minus 4/15. Why do the fifth evaluated for more equal 0 to 1? And then now we plug in one. We plug in zero maybe some tracks. And this gives us the lovely number of 16 route to over 15 minus 8/15. And so that's our expected value.

Here in this problem we are told to refer back To Exercise 4.36. Now it's important to remember that on that exercise we found that the expected value of acts Was there No 1? Right. And the variance of X was equal to two. Mhm. Now we're going to use this in order to find the given values here. Now we're told that Z is five X plus three. Uh huh. And so they expected values the is the expected value of five X plus three Which is five times the expected value. That's plus three. Yeah, just five times one plus three. Which is agents are expected value. The state the variance was the There's a variance of five, they're supposed three which is five squared times the variance of Acts, You know the variance of access to and so this is 25 times two which is 15. So the variance of the that's 50.

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.


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