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}}
another way of thinking about this is, what is tangent? sin over cosine. so there needs to be some argument such that sin(x)/cos(x) =√3. So look at the unit circle. what arguments involve the root of 3? π/6 does, but tan of π/6=sin(π/6)/cos(π/6)=1/√3. What else does? π/3. now what do you get?
well, the angle pi/3 gives you √3/2, correct? And you're multiplying on the outside by 2, right?
what's cos of the same angle?
you're using a different way to make your point. So you need angles to do it in the polar form.
If you memorize 3 or 4 pairs from the unit ciricle, I think, you...
several of my professors have said n=30 is a number that's been around for a while and is used as a rule of thumb, but you don't have to follow it super strictly. How skewed is the data? If it's not horribly skewed, if it's even vaguely bell-shaped, I'd imagine n=20 is acceptable. But please...
okay, so my first thought is that the joint then should be
\sum_{x_{2}=0}^{\infty}\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}
when I let p=.5, this term does indeed sum to 1, so that's looking alright.
oh, I think I maybe figured this out. So I should have (summing over values in the domain...
Homework Statement
Suppose that X1 and X2 have the joint pmf
f(x_{1},x_{2})=p^{2}q^{x_{2}},x_{1}=0,1,2,...,x_{2},x_{2}=0,1,2,...
with
0<p<1,q=1-p
Homework Equations
The Attempt at a Solution
I'm confused because the expression doesn't have x_1 in it. So usually, if I want to...
okay, so for the first one, can you construct an open ball around any point in A that is contained in A? If you look at Y on a cartesian plane, you have all the points (x,y) where x>0, and y>=0. if you look at A, you have all the points where x>0, and y>0. for every point a=(xo,y0) in A, is...
whenever i have a problem like this, i like to move my constants out front and rewrite things in fractions as negative exponents. in your case, i would rewrite the function as
ps/(1-qs)=p*(s)(1-qs)^-1. then, ignore the p for now, since you can just multiply it out, and focus on find the...
I'm guessing off your last reply that I'm not in fact allowed to do what I did, and I think I see the reason why: I switched the limits, but not the variables; so I first should have integrated f(t)dx as xf(t), which, evaluated from the proper limits should go to infinity minus negative infinity...
So I get that I can rewrite F(x+c)-F(x) as ∫xx+cf(x)dx....
is it alright for me to change the order of integration? Because then i would have
\intop_{-\infty}^{\infty}\int_{y}^{y+c}f(t)dtdx=\int_{y}^{y+c}\int_{-\infty}^{\infty}f(t)dtdx=\int_{y}^{y+c}1\cdot dx=(y+c)-y=c
But I'm not sure if...
Thank you for your reply. I'm still a bit lost. Is that integral a special form I should recognize? I'm still don't understand how I can say anything about what that equals without knowing what F(x) is. All I know is that it's non-negative and bounded by 0,1.
It took me a little while to...
Homework Statement
For any cdf F(x) of a continuous random variable, show that
\int_{-\infty}^{\infty}[F(x+b)-F(x+a)]dx=b-a
for any a<b.
Homework Equations
The Attempt at a Solution
Not really sure where to begin. I figure I can split the integrals and do u subs, and (after...