Probability Mass Function and Marginal Probaility

[EngWiPy]

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 $$\sum_yp_{X,Y}(x,Y=y)$$ ?

 

[EngWiPy]

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
$$P_X(1) = P_{XY}(1,1) + P_{XY}(1,2)$$

 

[EngWiPy]

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 $y^2 = b$.

 

[EngWiPy]

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.

 

[EngWiPy]

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

 

[EngWiPy]

Right

Thank you so much.