Recent content by mrkb80

Point taken. I'm going down this path now: F_{A+B+C} = P(A+B+C \le x) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{x-b-c} f_A(a) f_B(b) f_C(c) da db dc = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F_A(x-b-c) f_B(b) f_C(c) db dc Not sure where to go next...

Homework Statement Three yearly losses. First: Exponential Second & Third: Weibull Losses are independent. Find the 95% VaR of the min loss Homework Equations The Attempt at a Solution My first thought was: Let L be total loss, A be first Loss, B be second loss, C be third...

Homework Statement Assume regression model y_i = \alpha + \beta x_i + \epsilon_i with E[\epsilon_i] = 0, E[\epsilon^2] = \sigma^2, E[\epsilon_i \epsilon_j] = 0 where i \ne j. Suppose that we are given data in deviations from sample means. If we regress (y_i-\bar{y}) on (x_i-\bar{x}) without a...

cool. thanks again.

Homework Statement Let X_1,...,X_n be a random sample from a poisson distribution with mean \lambda Find the MLE of \lambda^2 + 1 Homework Equations The Attempt at a Solution I found \hat{\lambda}=\bar{x} Can I just square it and add 1 and solve for lambda hat? If not I have no idea...

Thanks for the reply. It just felt wrong to me because it was so far from the MLE estimate, but I know that can happen.

Homework Statement Let X_1,...,X_n be iid with pdf f(x;\theta) = \theta x^{\theta-1} , 0 \le x \le 1 , 0 < \theta < \infty Find an estimator for \theta by method of momentsHomework Equations The Attempt at a Solution I know I need to align the first moment of the beta distribution with the...

You're right. I see my mistake: f_X(x)=2e^{-x} makes much more sense.

I thought I understood it, but something is not correct. just working on the denominator, I get f_X(x)=-2(e^{-(x+y)}-e^{-x}) and then if I try to integrate that I get \int_0^1 f_X(x) dx = 2e^{-y-1}+2e^{-y}+e^{-1}+1 What am I missing here?

Great. Thanks for the help again.

many thanks, by the way.

I'm thinking that would be something like this: P(A) = P(Y<1) P(B) = P(X<1)=\int_0^1 f_X(x) dx so then P(A \cap B) =P(Y<1 \cap X<1) = \int_0^1 \int_0^y f_{X,Y}(x,y) dxdy and then \dfrac{\int_0^1 \int_0^y f_{X,Y}(x,y) dxdy}{\int_0^1 f_X(x) dx} or am I still not understanding?

Homework Statement let f_{X,Y}(x,y)=2e^{-(x+y)} for 0 \le x \le y and y \ge 0 \\ find P(Y<1 | X < 1) Homework Equations f(X=x | y=y) = \dfrac{f_{X,Y}(x,y)}{f_y(y)} The Attempt at a Solution P(Y<1 | X<1) = \int_0^1 \dfrac{f_{X,Y}(x,y)}{\int_0^1 f_X(x) dx} dy before I...

good point. I think I see my mistake(s): E[X] + E[Y]=\int_0^{\infty} x \lambda e^{-\lambda x} dx + \int_0^{\infty} y \lambda e^{-\lambda y} dy = - x \dfrac{1}{\lambda} e^{-\lambda x} |_0^{\infty} - \int_0^{\infty} e^{- \lambda x } dx - y \dfrac{1}{\lambda} e^{-\lambda y} |_0^{\infty} -...

Homework Statement Suppose that f_{X,Y}(x,y)=\lambda^2e^{-\lambda(x+y)},0\le x,0\le y find E[X+Y]Homework Equations The Attempt at a Solution I just want to double check I didn't make a mistake: E[X+Y]=E[X]+E[Y]=\int_0^{\infty} x{\lambda} e^{-\lambda x} dx + \int_0^{\infty} y{\lambda}...