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.
Homework Statement
Homework Equations
The Attempt at a Solution
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...
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?
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.
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}}