if the joint PMF is p(x,y)=k(x²+y²)

find the PMF 3X-2Y

is this given by substituting x=3X

y=-2Y

in which case it becomes:

p(3x,-2y)=k(9x²+4y²)

i am not sure if this is correct:

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- Aug 1st 2010, 10:28 PMulysses123joint Probability mass function
if the joint PMF is p(x,y)=k(x²+y²)

find the PMF 3X-2Y

is this given by substituting x=3X

y=-2Y

in which case it becomes:

p(3x,-2y)=k(9x²+4y²)

i am not sure if this is correct: - Aug 7th 2010, 09:34 AMSpringFan25
Your answer is not correct. You can see that is wrong by trying to use it to answer the question "what is the probability that 3x-2y = 5". There is no way to input this information into your function to get an answer.

Starting over, define a single variable Z = 3X - 2Y

You want the PMF of Z

$\displaystyle P(Z=z) = P(3X - 2Y = z) = P(Y=\frac{z+3X}{2})$

ie, to get a particular value of Z we cna have any value of X, provided that Y = (z+3x}/2

$\displaystyle P(Z=z) = \sum_x k \left[ x^2 + \left( \frac{z+3x}{2} \right)^2 \right]$

The last step is called the method of convolutions, which hopefully you have seen before if you were set this problem. - Aug 7th 2010, 09:07 PMmatheagle
I don't even know if the underlying joint distribution is discrete or continuous.

Nor do I know the support.