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

**Mathsdog** · April 13th 2012, 12:20 AM

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

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

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

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

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

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

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

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

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

Thanks in advance. MD

**ragnar** · April 13th 2012, 03:01 AM

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.

**Mathsdog** · April 13th 2012, 04:57 AM

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

**harish21** · April 13th 2012, 06:53 AM

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

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

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

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

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

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

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

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

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

Thanks in advance. MD

you can express the fraction as:

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

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

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

so your equation becomes:

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

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

see if you can go from here

**Mathsdog** · April 20th 2012, 12:35 AM

Hi,
Thanks for your previous comments. Thats v helpful.

As I see it I need to show that

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

right? Then the term containing x can come out of the sum and form the binomial, which gives $P_{X}(x)$ , 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.

**harish21** · April 20th 2012, 04:50 PM

Originally Posted by Mathsdog

Hi,
Thanks for your previous comments. Thats v helpful.

As I see it I need to show that

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

right? Then the term containing x can come out of the sum and form the binomial, which gives $P_{X}(x)$ , 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, $\left(1/6 \right)^{4-x-y} \neq \left(1/2 \right)^{4-x} \left(2/3 \right)^{4-x-y}$

The last step I had mentioned gives:

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

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

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

so you have:

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

the summation part now becomes 1. Why?

**Mathsdog** · April 23rd 2012, 11:15 PM

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

**harish21** · April 24th 2012, 04:32 PM

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

**Mathsdog** · April 24th 2012, 11:05 PM

Quite so. Thanks again. MD

**Mathsdog** · April 25th 2012, 11:43 PM

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

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

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

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

with the same conditions as above, right?

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

so,

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

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

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

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

**harish21** · April 27th 2012, 02:59 PM

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

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

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

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

with the same conditions as above, right?

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

so,

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

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

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

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:

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

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

pull out the non "z" terms

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

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

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

$\left(\frac{1}{6}\right)^{z} \left(\frac{5}{12} \times \frac{3}{5}\right)^{3-x-y-z}$
$=\left(\frac{1}{6}\right)^{z} \left(\frac{5}{12}\right)^{3-x-y-z}\left(\frac{3}{5}\right)^{3-x-y-z}$
$=\left(\frac{1}{6}\right)^{z} \left(\frac{5}{12}\right)^{3-x-y}\left(\frac{5}{12}\right)^{-z}\left(\frac{3}{5}\right)^{3-x-y-z}$
$=\left(\dfrac{(\frac{1}{6})^z}{(\frac{5}{12})^z} \right)\left(\frac{5}{12}\right)^{3-x-y}\left(\frac{3}{5}\right)^{3-x-y-z}$
$\left(\frac{2}{5}\right)^z\left(\frac{5}{12}\right )^{3-x-y}\left(\frac{3}{5}\right)^{3-x-y-z}$

So,

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

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

**Mathsdog** · April 28th 2012, 09:21 AM

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

**harish21** · April 28th 2012, 09:29 AM

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).

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