Because X and Y are independent. f_{y+3} (z|3) gives you the probability of a certain Y+3 given a certain X, but actually the value X takes doesn't influence Y, so you can just drop it.
You don’t always need two r.v’s to be independent to use the multiplication formula.independence does not always imply the marginals multiple, independence implies cov is 0 but opposite for cov isn’t true; and fy|x is same as fy and vice versa. Note that fy|x*fx(x)=f(x;y) if you have no information that x and y are ind or not. Now if you do, and they are ind, then the condition on y drops but nothing changes. So again, you don’t need x,y to be independent for the multiplication rule. Clearly z is dependent on x in this case. fz|x*fx(x)=f(x;z)
Can I get the full solution of the following question by today The Safe Grad Committe at a high school is selling raffle tickets on a Christmas Basket filled with gifts and gift cards.The prize is valued at $1200,the Committee has decided to sell 500 tickets only.What is the expected value of a ticket?If the students decide to sell tickets on three monetary prizes -one valued $1500 and two valued $ 500 each,what is the expected value of ticket now?
The expectation of a payout is equal to the sum of the weighted probability of each outcome - A) Since we have 1/500 chance of getting $1200 and 499/500 chance of getting $0, our expected value (what we would get on average, buying one ticket per year, every year) is 1200 * 1/500 + 0 * 499/500 which is equal to $2.40. Notice that this is the same amount that the committee should charge per ticket in order to break even - assuming all 500 tickets are sold. B) if we add up 1500 * 1/500 + 500 * 1/500 + 500 * 1/500 + 0 * 1/500, it can be seen plainly that we are getting 1/500 * (1500 + 500 + 500) overall. In other words, since our outcomes are equally likely, we can just sum together the different amounts and divide by 500... so our final answer is 2500 / 500 or $5. It should be observed that the decision of the committee to give one prize of a large amount or several smaller prizes doesn't effect the expected value of a ticket. Only when we increase the value of the total prize money does the expected value of a ticket increase.
Extremely easy to follow, thank you.
Oh a nice way of verifying! Thank you!!
this really clarified for me how to explicitly compute it, i was stuck for a while.
truly help me a lot
Helpful, thanks you!
why f_{X+b}(x)=f_X(x-b)？（3:56) i can't understand.
why f_{y+3} (z|3) = f_{y+3}(z) ? (3:10)
Because X and Y are independent. f_{y+3} (z|3) gives you the probability of a certain Y+3 given a certain X, but actually the value X takes doesn't influence Y, so you can just drop it.
How do you know that X and Z are independent? (I'm asking because they must be independent in order to use the multiplication rule)
they must be dependent because once you know little ' x ' you can know little bit about what 'z' might be
X and Y need to be independent
They're not independent. He didn't write f(X=x, Z=z) = f(X=x).f(Z=z) but f(X=x).f(Z=z | X=x).
You don’t always need two r.v’s to be independent to use the multiplication formula.independence does not always imply the marginals multiple, independence implies cov is 0 but opposite for cov isn’t true; and fy|x is same as fy and vice versa. Note that
fy|x*fx(x)=f(x;y) if you have no information that x and y are ind or not. Now if you do, and they are ind, then the condition on y drops but nothing changes. So again, you don’t need x,y to be independent for the multiplication rule. Clearly z is dependent on x in this case. fz|x*fx(x)=f(x;z)
You have to Let's Suppose
Can I get the full solution of the following question by today
The Safe Grad Committe at a high school is selling raffle tickets on a Christmas Basket filled with gifts and gift cards.The prize is valued at $1200,the Committee has decided to sell 500 tickets only.What is the expected value of a ticket?If the students decide to sell tickets on three monetary prizes -one valued $1500 and two valued $ 500 each,what is the expected value of ticket now?
The expectation of a payout is equal to the sum of the weighted probability of each outcome -
A) Since we have 1/500 chance of getting $1200 and 499/500 chance of getting $0, our expected value (what we would get on average, buying one ticket per year, every year) is 1200 * 1/500 + 0 * 499/500 which is equal to $2.40. Notice that this is the same amount that the committee should charge per ticket in order to break even - assuming all 500 tickets are sold.
B) if we add up 1500 * 1/500 + 500 * 1/500 + 500 * 1/500 + 0 * 1/500, it can be seen plainly that we are getting 1/500 * (1500 + 500 + 500) overall. In other words, since our outcomes are equally likely, we can just sum together the different amounts and divide by 500... so our final answer is 2500 / 500 or $5. It should be observed that the decision of the committee to give one prize of a large amount or several smaller prizes doesn't effect the expected value of a ticket. Only when we increase the value of the total prize money does the expected value of a ticket increase.