Rank-n Probability: Find Probability of x at k Position

[hugomcp]

Hello
Let y1,..yn be drawn from a normal distribution width known parameters.
Let x be drawn from another normal distribution with known parameters.
If the set {y1,...,yn, x} is ordered, what is the probability that x appears in the "k" position?

 

[mathman]

It depends very much on the parameters of these distributions. You need to be a little more specific. The simplest example is all having the same distribution, then the probability is the same for all positions.

 

[hugomcp]

It depends very much on the parameters of these distributions. You need to be a little more specific. The simplest example is all having the same distribution, then the probability is the same for all positions.

I know that the probability depends on the parameters of both distributions. If both have the same distribution, it will give an uniform. If x has a higher mean, it will give an exponential distribution.

I am looking for a formula that, taking into account the gaussian parameters of both distributions, computes the desired probability.

 

[ych22]

P(X appears in kth position) where k is an element from {0,1,2,...,n}
= P(X higher than k number of Y_i)
= (n C k) [P(X> Y_i)]^k[P(X< Y_i)]^n-k

 

[hugomcp]

P(X appears in kth position) where k is an element from {0,1,2,...,n}
= P(X higher than k number of Y_i)
= (n C k) [P(X> Y_i)]^k[P(X< Y_i)]^n-k

Thanks for your answer, but it may not be that simple. If that formula was true, then such probability for distributions with the sama parameters will give a normal, instead of a uniform, which I actually confirmed by simulation

 

[ych22]

Have you found the answer then?

 

[bpet]

P(X appears in kth position) where k is an element from {0,1,2,...,n}
= P(X higher than k number of Y_i)
= (n C k) [P(X> Y_i)]^k[P(X< Y_i)]^n-k

Not quite correct - the events {X>Y_i} are only conditionally independent, however conditioning on X will get a similar expression

P(X higher than k number of Y_i)
$$=\int_{-\infty}^{\infty}f(x)(^nC_k)F(x)^k(1-F(x))^{n-k}dx$$
where F is the cdf of the Y_i and f is the pdf of X, which reduces to 1/(n+1) when X has the same distribution as Y.