Probability density function ?

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Homework Statement


Suppose X selects an integer from the set S = {0,1,...,9} and Y selects an integer from {0,...,x^2}. Find:
(a) f(x,y) [joint prob density func]
(b) fY(y) [marginal for Y]
(c) Probability (Y <= 10 | X = 5)
(d) Probability (Y <= 10 | X <= 5)

Homework Equations


The Attempt at a Solution


I'm confused how to start this problem (finding the j.p.d.f.). My teacher's notes are kind of all over the place so this is what I attempted to put together:

(a) f(x,y) = Prob(X=x, Y=y) = f(y|x)f(x)
f(x) = 1/10
f(x,y) = 1/(x2+1) * (1/10)

(b) I'm guessing.. f(y) = Ʃ [ 1/(x^2+1) * (1/10) ]
I'm confused about what the summation is over. In the book it says "probability distribution h(y) of Y alone is obtained by summing f(x,y) over values of X". So does this mean f(y) = Ʃ [ 1/(x^2+1) * (1/10) ] from x=0 to x=9? Or is it from x=y0 (some fixed y) to x=9? If it is the latter case, how do I go about solving this?

Sry, i don't know if I mentioned it but (b) is supposed to be the marginal for Y (and Y alone)
 
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First you don't give a probability function for X or Y alone. Should we assume uniform probability?

You say "Suppose X selects an integer from the set S = {0,1,...,9} and Y selects an integer from {0,...,x^2}." Do you mean Suppose X selects an integer, x, from the set S = {0,1,...,9} and Y selects an integer from {0,...,x^2}? If not, what is x?
 
Sorry, the full problem says X uniformly selects an integer, x, from S={1,...,9}, and then Y uniformly selects an integer from {0,...,x^2}

Also, once I have (b), how do I apply that to (c) and (d)?
 
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f_y(y) will be a function of both x and y (but treats x as a constant parameter). Set x= 5 in (c). For (d), set x= 0, 1, 2, 3, 4, and 5 and sum.
 
Thanks for your help... I'm starting to see it now somewhat. One last question: is my fY(y) = (1/10) Ʃ (1/(x2+1)) from x=y0 to x=9 correct?

I got
(c) 42.3%
(d) ?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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