Recent content by phiiota

Okay, that makes sense. Thank you.

or did you want a specific case? say X~U(0,1), and I want to know say Y=X-1. then I can say g-1(y)=y-1., and d/dy g-1(y)=-y2, so then my distribution of Y would be f(g-1(y))|d/dy g-1(y)|=y-2. I get that if we didn't take the absolute value, then this function would be negative... But...

well, if say X distributed as f(X), and say i have Y=g(X), and I want to know the distribution of Y, in a simple case I can say that Y~f(g-1(y))|d/dy g-1(y)|. I don't know why we'd always want the absolute value, as opposed to the derivative in general.

this isn't really homework, but I was just wondering if someone could offer an intuitive reason as to why when random variables are transformed, we use absolute values of derivative of those functions, as opposed to the functions themselves?

Awesome, thanks. This concept took me a little while to get, but I think I got it. It's not very hard, but my book is kind of vague.

the fact that P(A) + P(B) > 1 tells you there has to be some overlap, right? So you're looking at kind of a best case/worst case scenario. How small could the overlap be? How large could the overlap be?

well, if it's conditional, and it's dependent on x, shouldn't your outcome be a function of x?

regarding your second question, just something to think about: what is P(A Union B)? what does this tell you about P(A intersect B)?

Homework Statement Suppose the distribution of X2 conditional on X1=x1 is N(x1,x12), and that the marginal distribution of X1 is U(0,1). Find the mean and variance of X2. Homework Equations Theorem: E(X_{2})=E_{1}(E_{2|1}(X_{2}|X_{1}))...

I could be wrong on this, so take this with a grain of salt. Apply the definition of expected value. It looks like you could rewrite your term as x!/(x-12)! Does this help? Then when you take the sum, the x! should cancel, and you'll be left with (x-12)! on the bottom.

when you have functions like this, sometimes it's easier to write out your systems in general terms. rewrite your equation as f(x)/g(x), or f(x)g(x)^-1, if you prefer, where f(x)=x and g(x)=√(3x+6). so now find the derivative of this. do it piece by piece until you have your solution.

what is the first thing you think of when you see "n+1"?

Thanks. As far as the latex goes, I usually type it in an editor, and then paste the code in here.

okay, so finally, we have that f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{i}=p^{2}\sum_{i=x_{1}}^{\infty}q^{i} so \sum_{i=0}^{\infty}q^{i}=\dfrac{1}{1-(1-p)}=\dfrac{1}{p}. \sum_{i=0}^{x_{1}}q^{i}=\dfrac{1-q^{x_{1}-1}}{1-q}=\dfrac{1-q^{x_{1}-1}}{p}, so taking the difference, we have...

Okay, so looking at a picture, I know that P(x1,x2)=0 whenever x1>x2 (and x1,x2>0). So I have kind of a triangular set up. So looking at these values, then, I'm thinking that my marginal for x should be f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{x_{1}}