CDF and PDF of order statistics

[EngWiPy]

Hi,

I have K i.i.d. exponentially distributed random variables with mean unity. I need to find the CDF and PDF of the summation of the largest two random variables. How can I do that? The problem in this case is that the combinations are not independent.

Thanks in advance

 

[AKG]

EDIT: Oops, sorry, when I said it was clear that the $X_i+X_j$ are clearly iid I was mistaken. I'm not sure that they're not independent, but if they are independent, it certainly isn't clear.

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The CDF of this random variable, let's call it $X$, is given by:

$$a \mapsto P(X < a)$$

What's the probability that the sum of the largest two of your exponentials is less than $a$? Well, the sum of the largest two is less than $a$ iff the sum of every two is less than $a$. In other words, if we let $X_1, \dots, X_K$ be your exponentials, then:

$$P(X < a) = P(X_1 + X_2 < a\mbox{ and }X_1 + X_3 < a\mbox{ and } \dots \mbox{ and }X_{K-1} + X_K < a)$$

Since the $X_i$ are iid, [STRIKE]it's not hard to see that the $X_i+X_j$ are iid[/STRIKE]. As such, we can rewrite the above:

$$P(X < a) = \prod_{1\leq i<j\leq K}P(X_i + X_j < a) = P(X_1 + X_2 < a)^{{K\choose 2}}$$

$a \mapsto P(X_1 + X_2 < a)$ is the CDF of a $\Gamma (2,1)$ distributed random variable.

 

[EngWiPy]

EDIT: Oops, sorry, when I said it was clear that the $X_i+X_j$ are clearly iid I was mistaken. I'm not sure that they're not independent, but if they are independent, it certainly isn't clear.

--------

The CDF of this random variable, let's call it $X$, is given by:

$$a \mapsto P(X < a)$$

What's the probability that the sum of the largest two of your exponentials is less than $a$? Well, the sum of the largest two is less than $a$ iff the sum of every two is less than $a$. In other words, if we let $X_1, \dots, X_K$ be your exponentials, then:

$$P(X < a) = P(X_1 + X_2 < a\mbox{ and }X_1 + X_3 < a\mbox{ and } \dots \mbox{ and }X_{K-1} + X_K < a)$$

Since the $X_i$ are iid, [STRIKE]it's not hard to see that the $X_i+X_j$ are iid[/STRIKE]. As such, we can rewrite the above:

$$P(X < a) = \prod_{1\leq i<j\leq K}P(X_i + X_j < a) = P(X_1 + X_2 < a)^{{K\choose 2}}$$

$a \mapsto P(X_1 + X_2 < a)$ is the CDF of a $\Gamma (2,1)$ distributed random variable.

Interesting! But are the events X1+X2<a and X1+X3<a are independent?

 

[chiro]

Hi,

I have K i.i.d. exponentially distributed random variables with mean unity. I need to find the CDF and PDF of the summation of the largest two random variables. How can I do that? The problem in this case is that the combinations are not independent.

Thanks in advance

Hey S_David.

The first thing is to get the distribution for the largest and second largest distribution. This can be done with order statistics.

Once you have these distributions, then if they are of the same type you can use MGF's to find the result type of adding the two (usually this is a good idea because in many situations adding to distributions that are i.i.d results in the same distribution with different parameters).

If the type is complex, use the convolution theorem to get the CDF and hence the PDF of the sum of the two variables. These variables have to be independent, but not necessarily identical distributed.

For convolution algorithm and more depth:

http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf

Formula for order statistic given a known PDF can be found here:

http://www.encyclopediaofmath.org/index.php/Order_statistic

 

[haruspex]

CDF of highest of n = Fⁿ(x)
2nd highest < x if either highest < x or exactly one (any of the n) > x:
CDF of 2nd highest = Fⁿ(x) + (1-Fⁿ(x))*n*F^n-1(x)
= n*F^n-1(x) - (n-1)*Fⁿ(x)
CDF of sum of highest two = ∫_y(n*F^n-1(x-y) - (n-1)*Fⁿ(x-y)).dFⁿ(y)
= ∫_y(n*F^n-1(x-y) - (n-1)*Fⁿ(x-y))*n*F^n-1(y).dF(y)

 

[haruspex]

CDF of highest of n = Fⁿ(x)
2nd highest < x if either highest < x or exactly one (any of the n) > x:
CDF of 2nd highest = Fⁿ(x) + (1-Fⁿ(x))*n*F^n-1(x)
= n*F^n-1(x) - (n-1)*Fⁿ(x)
CDF of sum of highest two = ∫_y(n*F^n-1(x-y) - (n-1)*Fⁿ(x-y)).dFⁿ(y)
= ∫_y(n*F^n-1(x-y) - (n-1)*Fⁿ(x-y))*n*F^n-1(y).dF(y)

On second thoughts, still an independence problem there.
Try 2:
P[2nd highest < x | highest = y > x] = (F(x)/F(y))^n-1
P[2nd highest < x | highest = y < x] = 1
CDF of sum of highest two = Fⁿ(x/2) + ∫_y>x/2(F(x-y)/F(y))^n-1.dFⁿ(y)
= Fⁿ(x/2) + n∫_y>x/2F(x-y)^n-1.dF(y)
Looks nicer at least.