A question about multinomial distribution

[Artusartos]

Let $X_1, ... , X_5$ be a joint multinomial with $n=15, p_1=.1, p_2=.15, p_3=.2, p_4=.24, p_5=.31$

What is the conditional distribution of $X_1, X_2, X_4, X_5$, given $X_1=3$


My answer:

Since $p(x_1, x_2, x_4, x_5 | x_3=3) = \frac{(15!) (1^{x_1}) (.1^{x_2}) (.15^{3}) (.2^{x_4}) (.31^{x_5})}{x_1! x_2! 3! x_4! x_5!}$

Do you think my answer is correct?

Thanks in advance.

 

[Ray Vickson]

Let $X_1, ... , X_5$ be a joint multinomial with $n=15, p_1=.1, p_2=.15, p_3=.2, p_4=.24, p_5=.31$

What is the conditional distribution of $X_1, X_2, X_4, X_5$, given $X_1=3$


My answer:

Since $p(x_1, x_2, x_4, x_5 | x_3=3) = \frac{(15!) (1^{x_1}) (.1^{x_2}) (.15^{3}) (.2^{x_4}) (.31^{x_5})}{x_1! x_2! 3! x_4! x_5!}$

Do you think my answer is correct?

Thanks in advance.

How did you arrive at your answer? If you go through the steps *carefully* you will be able to see for yourself whether the result is correct. Don't just write things down---go through the details,

RGV

 

[Artusartos]

How did you arrive at your answer? If you go through the steps *carefully* you will be able to see for yourself whether the result is correct. Don't just write things down---go through the details,

RGV

Since $p(x_1, x_2, x_4, x_5 | x_3=3) = p(x_1, x_2, x_3, x_4, x_5)/p(x_3)$...

Do you mean that I have to find $p(x_3)$ so I can divide by it? In order to do that I have to do integration, right? But its a bit confusing since they didn't tell us the boundaries, so how can I do the integration?

 

[Ray Vickson]

Since $p(x_1, x_2, x_4, x_5 | x_3=3) = p(x_1, x_2, x_3, x_4, x_5)/p(x_3)$...

Do you mean that I have to find $p(x_3)$ so I can divide by it? In order to do that I have to do integration, right? But its a bit confusing since they didn't tell us the boundaries, so how can I do the integration?

Yes, you need to find P{X3 = x3}.

The multinomial distribution is a DISCRETE distribution, with x_i = 0,1,2,...,n and some other restrictions. I don't know who is the "they" that did not tell you the boundaries, but material can be found in every textbook and on-line; see, eg., http://en.wikipedia.org/wiki/Multinomial_distribution .
There are no integrations involved, only summations.

You can save yourself a ton of work if you think about the meaning of the individual components in the distribution, and it might help to put it into some type of context. Suppose, for example, the 5 components correspond to apples, oranges, pears, bananas and grapes. If X_3 = number of pears, think about what it means to say that the number of pears is in the sample is 17 (for example) in a sample of 50 pieces of fruit (for example).

RGV

 

[Artusartos]

Yes, you need to find P{X3 = x3}.

The multinomial distribution is a DISCRETE distribution, with x_i = 0,1,2,...,n and some other restrictions. I don't know who is the "they" that did not tell you the boundaries, but material can be found in every textbook and on-line; see, eg., http://en.wikipedia.org/wiki/Multinomial_distribution .
There are no integrations involved, only summations.

You can save yourself a ton of work if you think about the meaning of the individual components in the distribution, and it might help to put it into some type of context. Suppose, for example, the 5 components correspond to apples, oranges, pears, bananas and grapes. If X_3 = number of pears, think about what it means to say that the number of pears is in the sample is 17 (for example) in a sample of 50 pieces of fruit (for example).

RGV

Oh...sorry, I don't know why I thought of it as continuous...