Distribution of X for an Urn with n Balls

[DenisaZsuZsa]

Homework Statement

You have an urn that contains n balls labeled with the natural numbers {1,2,3,...,n} and you extract n balls from the urn (with the condition that a ball may not be returned to the urn once drawn). You have to determine the distribution of X=(X₁,X₂,...,X_n), where X_k is the number of the ball from the k-th extraction.

Homework Equations

For Z a discrete random variable, that can take the values 1, ..., n

Probability Distribution
p(z)=P(Z=z)

Cumulative Distribution Function
F(z)=P(Z<z)

where z is in {1,...,n}

The Attempt at a Solution

There are n! possible cases.

X_k (k={1,...,n}) are discrete random variable. Before the k-th extraction in the urn there are n-k+1 balls left with the probability of occurrence = 1/(n-k+1) and X_k are discrete random variable:

1 2 ... n​

X₁: ( 1/n 1/n ... 1/n )


considering i₁ the number of the ball extracted on the first extraction, we have:

1 ... i₁-1 i₁ i₁+1 .. n​

X₂: ( 1/(n-1) ... 1/(n-1) 0 1/(n-1) .. 1/(n-1) )

that means that for i₁ the probability of occurrence = 0.
...

before the last extraction there is 1 ball left in urn, that obviously, has the probability of occurrence = 1.

I have no clue what to do next and I need it to demonstrate something for a bigger project. I do not ask you for the solution, I just need some hints.


P.S.: I hope you will understand what I wrote here, I've tried my best. Sorry if you don't, english is not my native language.

 

[HallsofIvy]

The probability that the first ball is labeled $x_1$ for any $x_1$ is 1/n. The probability that the second ball is labeled $x_2$ for any $x_2$ other than [itex]x_1[/tex] is 1/(n-1), etc. Thus the probability $P(X)= P(X_1, X_2, ..., X_n)$$= (1/n)(1/(n-1))(1/(n-2))..., (1/2)(1/1)= 1/n!$ as long as X is a permutation of (1, 2, 3,..., n) and 0 if it is not (that is, if the same integer occurs more than once in the "X"s).[/itex]

 

[DenisaZsuZsa]

The probability that the first ball is labeled $x_1$ for any $x_1$ is 1/n. The probability that the second ball is labeled $x_2$ for any $x_2$ other than [itex]x_1[/tex] is 1/(n-1), etc. Thus the probability $P(X)= P(X_1, X_2, ..., X_n)$$= (1/n)(1/(n-1))(1/(n-2))..., (1/2)(1/1)= 1/n!$ as long as X is a permutation of (1, 2, 3,..., n) and 0 if it is not (that is, if the same integer occurs more than once in the "X"s).[/itex]

$<br /> That is logic, cause it's: the number of favorable cases/number of posible cases => 1/n!<br /> Also you can't say P(X), you have to say P(X=value) or P(X<value), so the corect way is: P(X=(x₁,...x_n))=P(X₁=x₁, ...X_n=x_n)<br /> <br /> What I do not know is how to put the condition: X is a permutation of (1, 2, ...,n), meaning that I don't not how to express the dependecies between X₁ and X₂ , between X₂ and X₃ and so on...<br /> <br /> <br /> If I didn't have the condition: X is a permutation of (1, 2, ...,n), in other words if they where independent, I could say that E[X_i]=(n+1)/ 2 and Var(X_i)= (n-1)²/12, but they aren't independent.<br /> <br /> How can I calculate the mean and Var in this case?$