Question on Probability & Uniform Distribution.

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The discussion focuses on determining the distributions of the first and highest order statistics from a sample drawn from a continuous uniform distribution. The first order statistic (minimum) can be approached by considering the probability that all sampled values exceed a certain threshold, leading to the formula P(X_{(1)} > a) = 1 - (1 - F_X(a))^n. For the highest order statistic (maximum), the probability is derived from the independence of the sampled values, resulting in P(X_n ≤ a) = (F(a))^n. Participants clarify that the distribution of order statistics is not Binomial, emphasizing the need to apply the properties of independence and cumulative distribution functions. The conversation highlights the importance of understanding these statistical concepts for solving the posed problem effectively.
Sunil12
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Suppose a sample of random size N is taken from the continuous uniform(0, θ)
distribution, and N has a discrete distribution with p.m.f.

P (N = n) = 1/(n! (e − 1) ) for n = 1, 2, 3, . . . .

Determine the distribution of the
i) first order statistic (the minimum) of X1 , X2, . . . , XN .
ii) highest order statistic (the maximum) of X1, X2, . . ., XN .

Please help me to solve this problem.
 
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Sounds like homework. What have you tried? What do you know in general about the distribution of order statistics?
 
Generally the distribution of Xi in a order statistic is Binomial. Isn't it ? because given a value 'a' Xi is either Xi < a or Xi>=a. It is like a success failure.
 
To get you started, P[max<=x|N=n] = P[X1<=x,...,Xn<=x]=P[X1<=x]^n, then simply take the expectation wrt N.
 
Sunil12 said:
Generally the distribution of Xi in a order statistic is Binomial. Isn't it ? because given a value 'a' Xi is either Xi < a or Xi>=a. It is like a success failure.

Not quite. If you know that

<br /> P(X_n \le a)<br />

(X_n is the largest order statistic) then you know that ALL the other values are less than or equal to a, so

<br /> P(X_n \le a) = P(X1 \le a \text{ and } X2 \le a \text{ and } \cdots \text{ and } Xn \le a)<br /> = \left(P(X \le a)\right)^n = F(a)^n<br />

by independence. To work with the minimum start with

<br /> P(X_{(1)} &gt; a)<br />

and think about what it means for the smallest item in the sample to be larger than some value.
 
FX1(a) = 1 - P(X1 > a)

which will essentially be 1 - (1 - Fx(x))n

Right ?
 
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