Probability density function ?

In summary, the problem involves finding the joint probability density function for two variables, X and Y, where X is selected from the set S = {0,1,...,9} and Y is selected from {0,...,x^2}. The marginal probability density function for Y can be obtained by summing the joint probability density function over all values of X. To find the probability (Y <= 10 | X = 5), set X = 5 in the marginal probability density function for Y. To find the probability (Y <= 10 | X <= 5), sum the marginal probability density function for Y over values of X from 0 to 5.
  • #1
csc2iffy
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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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  • #2
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?
 
  • #3
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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  • #4
[itex]f_y(y)[/itex] 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.
 
  • #5
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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FAQ: Probability density function ?

What is a probability density function (PDF)?

A probability density function (PDF) is a mathematical function that describes the relative likelihood of different outcomes of a random variable. It is used to model continuous probability distributions.

2. How is a PDF different from a probability distribution function (PDF)?

A PDF is a function that describes the relative likelihood of different outcomes of a continuous random variable, while a probability distribution function (PDF) describes the probability of each possible outcome of a discrete random variable.

3. What is the relationship between a PDF and a cumulative distribution function (CDF)?

The PDF and CDF are related by the fundamental theorem of calculus. The PDF is the derivative of the CDF, which means that the CDF can be obtained by integrating the PDF over a given interval.

4. How is a PDF used in statistics and data analysis?

A PDF is used in statistics and data analysis to calculate the probability of a continuous random variable falling within a certain range or interval. It is also used to estimate the likelihood of a particular outcome occurring based on a given set of data.

5. Can a PDF have negative values?

No, a PDF must always have non-negative values. This is because the probability of an event cannot be negative. Therefore, the PDF must always be non-negative to ensure that the total probability of all possible outcomes is equal to 1.

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