Recent content by BookMark440

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    Moment generating function and expectation

    That link was very helpful. I completed the problem but I am a little uncertain about my strategy of finding the derivative by parts and the answer: X has p.d.f : p(j) = 2^(-j), j=1,2,3,... I computed the MGF for X = e^t/(2-e^t) Then E[X] = d/dt (e^t/(2-e^t)) [evaluated for t= 0] =...
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    Moment generating function and expectation

    Homework Statement Let X denote a random variable with the following probability mass function: P(j)= 2^(-j), j=1,2,3,... (a) Compute the moment generating function of X. (b) Use your answer to part (a) to compute the expectation of X. Homework Equations m.g.f of X is M (t) =...
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    How do I compute P(y>x) given f(x) and f(y|x)?

    Maybe I am looking for the wrong answer. Do I simply need to find the portion of f(x)=exp(-x) that are to the left of its intersection with f(x)=x ? This would mean I am looking for the intersection, which is exp(-x) = x, solving for x? That is: -x*log e = log x 0.4343 = -(log x)/x But what...
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    How to Compute Conditional Probability Without Deriving f(y)

    I understand that part. My problem (I think) is that I need to evaluate when f(y) has points above the line x=y and I do not know how to derive f(y). Does that make any sense?
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    How do I compute P(y>x) given f(x) and f(y|x)?

    Homework Statement Given f(x) = e^-x and f(y|x) = 1/x e^(-y/x). Three parts: (a) Compute density of (x,y), (b) Compute E(y) and (c) Compute P(y>x). Homework Equations f(x,y) = f(y|x)f(x) if f(x) = ve^(-vx), then E(x)=v^(-1) The Attempt at a Solution I'm stuck on a problem. I was...
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    How to Compute Conditional Probability Without Deriving f(y)

    I'm stuck on a problem. I was given f(x) and f(y|x) and was able to derive f(x,y). The second step of the problem is computing P[y>x]. I think I need to know f(y) to answer this problem but I can't figure out how to derive it. Or is there a way to compute P(y>x) given the info I know without...
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    The Cov(X,Y) on a square dart board

    Thanks. I did that and ended up getting zero for an answer, which always makes me think I made an error. In this case, I guess it just verifies that X,Y are independent.
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    The Cov(X,Y) on a square dart board

    A point (X, Y) is picked at random, uniformly from the square with corners at (0,0), (1,0), (0,1) and (1,1). Compute Cov{X, Y}. I think of this as darts thrown at a unit square dart board. Cov(X,Y) = E(XY) - E(X)E(Y). I compute that E(X)=E(Y)= 0.5 using the integral xf(x). But I...
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