Probability Mass Function and Marginal Probaility

[S_David]

Hi,

If I have a joint probability mass function p_{X,Y}(x,y), can we get the marginal probability mass functions p_X(x) and p_Y(y), without any knowledge of the conditional probability function of either of them, and the probability of each event? I mean, I know that:

p_X(x)=\sum_yp_{X,Y}(x,Y=y)=\sum_yp(x/Y=y)\text{Pr}\{y\}

But I just have p_{X,Y}(x,y). Can I?

Thanks

[Stephen Tashi]

Hi,

p_X(x)=\sum_yp_{X,Y}(x,Y=y)=\sum_yp(x/Y=y)\text{Pr}\{y\}

But I just have p_{X,Y}(x,y). Can I?

Thanks

If you know p_{X,Y}(x,y), what would stop you from computing \sum_yp_{X,Y}(x,Y=y) ?

[S_David]

If you know p_{X,Y}(x,y), what would stop you from computing \sum_yp_{X,Y}(x,Y=y) ?

How? Let us assume for the sake of the argument that p_{X,Y}(x,y) is 0.4 when x=y=1, and 0.6 when x=y=2. Now according to the equation you pointed to, we have:

p_X(x)=p_{X,Y}(x,Y=1)+p_{X,Y}(x,Y=2)

Ok? then what?

[Stephen Tashi]

If you "have" P_{XY}(x,y) then you know the value of things like P_{XY}(1,2) and P_{XY}(2,1) so there is no problem doing those sums.

For example, if the only possible values of the variables are 1 and 2, then
P_X(1) = P_{XY}(1,1) + P_{XY}(1,2)

[S_David]

If you "have" P_{XY}(x,y) then you know the value of things like P_{XY}(1,2) and P_{XY}(2,1) so there is no problem doing those sums.

For example, if the only possible values of the variables are 1 and 2, then
P_X(1) = P_{XY}(1,1) + P_{XY}(1,2)

Good. Now what if we need to find that joint p.m.f of X^2\text{ and }Y^2. That is, p_{X^2,Y^2}(x^2,y^2) from p_{X,Y}(x,y)?

Thanks for helping.

[Stephen Tashi]

Good. Now what if we need to find that joint p.m.f of X^2\text{ and }Y^2. That is, p_{X^2,Y^2}(x^2,y^2) from p_{X,Y}(x,y)?

Thanks for helping.

If g(r,s) is the joint p.m.f of (X^2,Y^2) , to find g(a,b) , you must sum p_{XY}(x,y) over all combinations of (x,y) that give x^2 = a and y^2 = b .

[S_David]

If g(r,s) is the joint p.m.f of (X^2,Y^2) , to find g(a,b) , you must sum p_{XY}(x,y) over all combinations of (x,y) that give x^2 = a and y^2 = b .

Ok, I am not getting the idea very well. Let me try to write an equation of this. Using your notation, we have:

g(a,b)=\sum_{(x,y)}p_{X,Y}(x:x^2=a,y:y^2=b)

where : means such that. So, we have the following choices of x and y that satisfy the equation:

x=\pm a \text{ and }y=\pm b

Am I right so far?

[Stephen Tashi]

You meant \sqrt{a} and \sqrt{b} , but yes, that's the general idea.

[S_David]

You meant \sqrt{a} and \sqrt{b} , but yes, that's the general idea.

Yes you are right, the square root of the values. So, for the example I gave previously, we have the following:

g(a,b)=\left\{\begin{array}{cc}0.4&a=b=1\\0.6&a=b=4\end{array}\right.

right?

[Stephen Tashi]

Right

[S_David]

Right

Thank you so much.