Marginal Probability

[kuahji]

Find the marginal probability functions of a) X and b) Y
f(x,y)=cxy for x=1,2,3 y=1,2,3 (discrete data)

The first thing I did was solve for c, which turned out to be 1/36. The book confirmed this.

Next I setup a table, for x=1 the total would be 6c, x=2 12c, x=3 18c.

So for the marginal probability for X, where x=1 I took 6 (1/36) 1/6, x=2 12(1/36)=1/3, & finally x=3 18(1/36)=1/2

so the marginal probability for X is as follows
f(x)= 1/6 for x=1
1/3 for x=2
1/2 for x=3

However, this is what the book has for an answer
f(x)= x/6 for x=1,2,3

So I'm kinda confused here. The book has a similar example & I did this problem exactly like the example, so where am I going wrong?

[Zone Ranger]

Find the marginal probability functions of a) X and b) Y
f(x,y)=cxy for x=1,2,3 y=1,2,3 (discrete data)

The first thing I did was solve for c, which turned out to be 1/36. The book confirmed this.

Next I setup a table, for x=1 the total would be 6c, x=2 12c, x=3 18c.

So for the marginal probability for X, where x=1 I took 6 (1/36) 1/6, x=2 12(1/36)=1/3, & finally x=3 18(1/36)=1/2

so the marginal probability for X is as follows
f(x)= 1/6 for x=1
1/3 for x=2
1/2 for x=3

However, this is what the book has for an answer
f(x)= x/6 for x=1,2,3

So I'm kinda confused here. The book has a similar example & I did this problem exactly like the example, so where am I going wrong?

The book is correct. You need to sum over y.

f_X(x)=\sum_{y=1}^3f_{X,Y}(x,y)=\sum_{y=1}^3\frac{ 1}{36}xy=\frac{1}{36}x(1+2+3)=\frac{x}{6}

Edit: I just noticed that you do have the correct answer. Think about it.

[kuahji]

Indeed, you are correct... :)