Recent content by phiiota

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    Intuitive reason absolute values are used for transformations in statistics?

    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...
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    Intuitive reason absolute values are used for transformations in statistics?

    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.
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    Intuitive reason absolute values are used for transformations in statistics?

    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...
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    Expected value and variance of a conditional pdf (I think I have it)

    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.
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    Using Chebyshev's Theorem (and another minor question)

    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?
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    Integration of random variables

    well, if it's conditional, and it's dependent on x, shouldn't your outcome be a function of x?
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    Using Chebyshev's Theorem (and another minor question)

    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)?
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    Expected value and variance of a conditional pdf (I think I have it)

    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}))...
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    Poisson distribution questions

    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.
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    Find dy of x/( sqrt (3x + 6) )

    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.
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    Finding convergence of a recursive sequence

    what is the first thing you think of when you see "n+1"?
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    Finding a marginal pmf

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

    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...
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    Finding a marginal pmf

    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}}
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