Marginal of a joint multinomial reduces to a binomial, but how?

Hi,

I'd be very grateful if someone can show me the nuts and bolts of how this works

Suppose that X and Y are discrete random variables with

$\displaystyle P_{X,Y}(x, y) = {4! \over x! y! (4-x-y)!} \left(\frac{1}{2}\right)^{x} \left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$

Where $\displaystyle 0 \leq x + y \leq 4$

Now, I know the marginals are binomials with n=4 and p = 1/2 for $\displaystyle P_{X}(x)$ and p = 1/3 for $\displaystyle P_{Y}(y)$.

So just takin X for now, as I see it the marginal for X, $\displaystyle P_{X}(x)$, should be

$\displaystyle P_{X}(x) = \sum_{y =0}^{4-x} {4! \over x! y! (4-x-y)!} \left(\frac{1}{2}\right)^{x} \left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$

But how does this, if it is correct, reduce to

$\displaystyle P_{X}(x) =\binom{4}{x}\left(\frac{1}{2}\right)^{x} \left(\frac{1}{2}\right)^{4-x}$

? I cant see it, and I'd love to.

Thanks in advance. MD

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Since the numbers are small, you could do this super-manually: Take your binomial and plug in, successively, 0,...,4. Then take your sum formula and successively plug in 0,...,4 for x, and see that upon summing over y in each case, you obtain the same result and therefore the functions are identical.

I'm not really sure of a more general technique that would be informative and handle cases of larger n.

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Hi, yep. Thanks for your interest, but its not really what I'm looking form. Yes, obviously if my sum is equivalent to the binomial then they should be numerically equivalent too. I'm really looking for an analytical equivalence. Sorry, I should have stated that more explicitly. Thanks again for the interest though. MD

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Quote:

Originally Posted by

**Mathsdog** Hi,

I'd be very grateful if someone can show me the nuts and bolts of how this works

Suppose that X and Y are discrete random variables with

$\displaystyle P_{X,Y}(x, y) = {4! \over x! y! (4-x-y)!} \left(\frac{1}{2}\right)^{x} \left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$

Where $\displaystyle 0 \leq x + y \leq 4$

Now, I know the marginals are binomials with n=4 and p = 1/2 for $\displaystyle P_{X}(x)$ and p = 1/3 for $\displaystyle P_{Y}(y)$.

So just takin X for now, as I see it the marginal for X, $\displaystyle P_{X}(x)$, should be

$\displaystyle P_{X}(x) = \sum_{y =0}^{4-x} {4! \over x! y! (4-x-y)!} \left(\frac{1}{2}\right)^{x} \left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$

But how does this, if it is correct, reduce to

$\displaystyle P_{X}(x) =\binom{4}{x}\left(\frac{1}{2}\right)^{x} \left(\frac{1}{2}\right)^{4-x}$

? I cant see it, and I'd love to.

Thanks in advance. MD

you can express the fraction as:

$\displaystyle {4! \over x! y! (4-x-y)!} = \dfrac{4!}{x! y! (4-x-y)!} \times \dfrac{(4-x)!}{(4-x)!} $

$\displaystyle =\dfrac{4!}{x!(4-x)!} \times \dfrac{(4-x)!}{y!(4-x-y)!}$

$\displaystyle =\dbinom{4}{x}\dbinom{4-x}{y}$

so your equation becomes:

$\displaystyle P_{X}(x) = \sum_{y =0}^{4-x}\dbinom{4}{x}\dbinom{4-x}{y} \left(\frac{1}{2}\right)^x \left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$

$\displaystyle P_{X}(x)=\dbinom{4}{x} \left(\frac{1}{2}\right)^x \sum_{y =0}^{4-x}\dbinom{4-x}{y}\left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$

see if you can go from here

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Hi,

Thanks for your previous comments. Thats v helpful.

As I see it I need to show that

$\displaystyle \left(1/6 \right)^{4-x-y} = \left(1/2 \right)^{4-x} \left(2/3 \right)^{4-x-y}$

right? Then the term containing x can come out of the sum and form the binomial, which gives $\displaystyle P_{X}(x)$, while the sum over y is equal to 1 and therefore can be forgotten about.However, I just can't derive the identity above.

Thanks in advance for any comments. MD.

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Quote:

Originally Posted by

**Mathsdog** Hi,

Thanks for your previous comments. Thats v helpful.

As I see it I need to show that

$\displaystyle \left(1/6 \right)^{4-x-y} = \left(1/2 \right)^{4-x} \left(2/3 \right)^{4-x-y}$

right? Then the term containing x can come out of the sum and form the binomial, which gives $\displaystyle P_{X}(x)$, while the sum over y is equal to 1 and therefore can be forgotten about.However, I just can't derive the identity above.

Thanks in advance for any comments. MD.

First of all, $\displaystyle \left(1/6 \right)^{4-x-y} \neq \left(1/2 \right)^{4-x} \left(2/3 \right)^{4-x-y}$

The last step I had mentioned gives:

$\displaystyle P_{X}(x)=\dbinom{4}{x} \left(\frac{1}{2}\right)^x \sum_{y =0}^{4-x}\dbinom{4-x}{y}\left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$

you can write $\displaystyle \left(\frac{1}{3}\right)^{y} \left(\frac{1}{6}\right)^{4-x-y}$ as:

$\displaystyle \left(\frac{1}{3}\right)^{y} \left(\frac{1}{2} \times \frac{1}{3}\right)^{4-x-y}$

$\displaystyle =\left(\frac{1}{3}\right)^{y} \left(\frac{1}{2}\right)^{4-x-y}\left(\frac{1}{3}\right)^{4-x-y}$

$\displaystyle =\left(\frac{1}{3}\right)^{y} \left(\frac{1}{2}\right)^{4-x}\left(\frac{1}{2}\right)^{-y}\left(\frac{1}{3}\right)^{4-x-y}$

$\displaystyle =\left(\dfrac{(\frac{1}{3})^y}{(\frac{1}{2})^y} \right)\left(\frac{1}{2}\right)^{4-x}\left(\frac{1}{3}\right)^{4-x-y}$

$\displaystyle \left(\frac{2}{3}\right)^y\left(\frac{1}{2}\right) ^{4-x}\left(\frac{1}{3}\right)^{4-x-y}$

so you have:

$\displaystyle P_{X}(x)=\dbinom{4}{x} \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{4-x}\sum_{y =0}^{4-x}\dbinom{4-x}{y}\left(\frac{2}{3}\right)^y\left(\frac{1}{3} \right)^{4-x-y}$

the summation part now becomes 1. Why?

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Thanks v much. That's really helpful. V nice too. The summation becomes 1 because this is a probability distribution (bernoulli with n = 4-x and p = 2/3) and therefore sums to one since the sum is over the whole range of outcomes. Is that how you might put it? Thanks again. MD

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Quote:

Originally Posted by

**Mathsdog** Thanks v much. That's really helpful. V nice too. The summation becomes 1 because this is a probability distribution (bernoulli with n = 4-x and p = 2/3) and therefore sums to one since the sum is over the whole range of outcomes. Is that how you might put it? Thanks again. MD

Binomial with n = 4-x and p = 2/3

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Quite so. Thanks again. MD

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Hi Harish21,

I wonder if I might bounce another one off you. This time it's three variables, summing one out. It should be similar to the previous one, but again I must be being even dumber than usual.

It is:

Given the joint: $\displaystyle p_{X, Y, Z}(x, y, z) = \frac{3!}{x!y!z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

Where $\displaystyle x, y, z = 0, 1, 2, 3$ and $\displaystyle 0 \leq x + y + z \leq 3$

Find the joint marginal $\displaystyle p_{X, Y}(x, y)$

Which, I think, is: $\displaystyle p_{X, Y}(x, y) = \sum_{z=0}^{3-x-y}\frac{3!}{x!y!z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

with the same conditions as above, right?

I did the same trick you did before with the factor I think

so,

$\displaystyle \frac{3!}{x!y!z!(3-x-y-z)!} \cdot \frac{(3- x- y)!}{(3- x- y)!} = \frac{3!}{x!y!(3-x-y)!} \cdot \frac{(3- x- y)!}{z!(3-x-y-z)!}$

Which gives the right factor for a joint multinomial for x and y, n=3, $\displaystyle p_{1} = 1/2$ and $\displaystyle p_{2} = 1/12$

i.e. $\displaystyle p_{X,Y}(x, y)=\frac{3!}{x!y!(3-x-y)!} \cdot (1/2)^x (1/12)^y (5/12)^{3-x-y}$

Which is what the sum over z should reduce to, right?

But, damn it, I can't seem to massage out this distribution from the sum over z in a similar way to before.

Thanks in advance for any help. BW, MD

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Quote:

Originally Posted by

**Mathsdog** Hi Harish21,

I wonder if I might bounce another one off you. This time it's three variables, summing one out. It should be similar to the previous one, but again I must be being even dumber than usual.

It is:

Given the joint: $\displaystyle p_{X, Y, Z}(x, y, z) = \frac{3!}{x!y!z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

Where $\displaystyle x, y, z = 0, 1, 2, 3$ and $\displaystyle 0 \leq x + y + z \leq 3$

Find the joint marginal $\displaystyle p_{X, Y}(x, y)$

Which, I think, is: $\displaystyle p_{X, Y}(x, y) = \sum_{z=0}^{3-x-y}\frac{3!}{x!y!z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

with the same conditions as above, right?

I did the same trick you did before with the factor I think

so,

$\displaystyle \frac{3!}{x!y!z!(3-x-y-z)!} \cdot \frac{(3- x- y)!}{(3- x- y)!} = \frac{3!}{x!y!(3-x-y)!} \cdot \frac{(3- x- y)!}{z!(3-x-y-z)!}$

Which gives the right factor for a joint multinomial for x and y, n=3, $\displaystyle p_{1} = 1/2$ and $\displaystyle p_{2} = 1/12$

i.e. $\displaystyle p_{X,Y}(x, y)=\frac{3!}{x!y!(3-x-y)!} \cdot (1/2)^x (1/12)^y (5/12)^{3-x-y}$

Which is what the sum over z should reduce to, right?

But, damn it, I can't seem to massage out this distribution from the sum over z in a similar way to before.

Thanks in advance for any help. BW, MD

Its pretty similar to the one above. SO you have:

$\displaystyle p_{X, Y}(x, y) = \sum_{z=0}^{3-x-y}\frac{3!}{x!y!z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

$\displaystyle = \sum_{z=0}^{3-x-y} \frac{3!}{x!y!(3-x-y)!} \cdot \frac{(3- x- y)!}{z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

pull out the non "z" terms

$\displaystyle = \frac{3!}{x!y!(3-x-y)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \sum_{z=0}^{3-x-y} \frac{(3- x- y)!}{z!(3-x-y-z)!}\left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

$\displaystyle = \frac{3!}{x!y!(3-x-y)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \sum_{z=0}^{3-x-y} \frac{(3-x-y)!}{z!(3-x-y-z)!}\left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

you can write $\displaystyle \left(\frac{1}{6}\right)^{z} \left(\frac{1}{4}\right)^{3-x-y-z}$ as:

$\displaystyle \left(\frac{1}{6}\right)^{z} \left(\frac{5}{12} \times \frac{3}{5}\right)^{3-x-y-z}$

$\displaystyle =\left(\frac{1}{6}\right)^{z} \left(\frac{5}{12}\right)^{3-x-y-z}\left(\frac{3}{5}\right)^{3-x-y-z}$

$\displaystyle =\left(\frac{1}{6}\right)^{z} \left(\frac{5}{12}\right)^{3-x-y}\left(\frac{5}{12}\right)^{-z}\left(\frac{3}{5}\right)^{3-x-y-z}$

$\displaystyle =\left(\dfrac{(\frac{1}{6})^z}{(\frac{5}{12})^z} \right)\left(\frac{5}{12}\right)^{3-x-y}\left(\frac{3}{5}\right)^{3-x-y-z}$

$\displaystyle \left(\frac{2}{5}\right)^z\left(\frac{5}{12}\right )^{3-x-y}\left(\frac{3}{5}\right)^{3-x-y-z}$

So,

$\displaystyle p_{X, Y, Z}(x, y, z)= \frac{3!}{x!y!(3-x-y)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left(\frac{5}{12}\right)^{3-x-y}\underbrace{\sum_{z=0}^{3-x-y} \frac{(3-x-y)!}{z!(3-x-y-z)!}\left( \frac{2}{5} \right)^z \left( \frac{3}{5} \right)^{3-x-y-z}}_{\text{This is a Binomial}(3-x-y, 2/5)}$

so then you have a binomial distribution summing up to 1.. and the remaining terms will give a multinomial distribution if you are familiar with it.

Multinomial distribution - Wikipedia, the free encyclopedia

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Hey man. That is really cool. I think I need to just practise a bit to get it down pat. You are pretty good at this stuff. . . what do you do if you dont mind my asking? I see you are a student. I am doing a neuroscience PhD.

Anyway, thanks again. Jon

Re: Marginal of a joint multinomial reduces to a binomial, but how?

Quote:

Originally Posted by

**Mathsdog** Hey man. That is really cool. I think I need to just practise a bit to get it down pat. You are pretty good at this stuff. . . what do you do if you dont mind my asking? I see you are a student. I am doing a neuroscience PhD.

Anyway, thanks again. Jon

Its all about practice. You'll get better at it. I am studying Biostatistics(MS).