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Given the pdf $f(x)=\begin{cases}2x&\text{0<x<1}\\0 & \text{otherwise}\end{cases}$. What is the probability that the sample median based on a random sample of size 3 drawn from the distribution with pdf f(x) exceeds 1/2?

Here, although I can calculate the value of the median by integrating f(x) from $0$ to median=m(say) and then equating it to $1/2$. The value for median came $1/\sqrt2$. But I don't know how to find the probability for the median now.

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    $\begingroup$ Please add the self-study tag. The population median that you found is a constant. Question asks for the sample median which is random. Do you know order statistics? $\endgroup$ Commented Sep 28, 2019 at 14:53
  • $\begingroup$ Yes I do know order statistics. $\endgroup$
    – Azka
    Commented Sep 28, 2019 at 15:15
  • $\begingroup$ That would be the 2nd value of the sample. $\endgroup$
    – Azka
    Commented Sep 28, 2019 at 15:59
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    $\begingroup$ Yes that would be $X_{(2)}$ $\endgroup$
    – Azka
    Commented Sep 28, 2019 at 16:30
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    $\begingroup$ So that means I need to calculate $P(X_{(2)}>1/2)$. $\endgroup$
    – Azka
    Commented Sep 28, 2019 at 18:30

2 Answers 2

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Alexander Pope wrote "A little learning is a dangerous thing, ..." but great learnings (as in BruceET's answer and in the link posted by whuber) can create unneeded diversions in solving little problems.

We have three independent random variables $X_1, X_2, X_3$ for which we readily can compute that $P(X_i < \frac 12) = \frac 14$ and $P(X_i > \frac 12) = \frac 34$. We are asked for the probability that at least two of the $X_i$ exceed $\frac 12$. Well, the probability that all three exceed $\frac 12$ is $\left(\frac 34\right)^3 = \frac{27}{64}$ while the probability that exactly two of the $X_i$ exceed $\frac 34$ is $3\times \frac 14\times \left(\frac 34\right)^2 = \frac{27}{64}$, making the desired probability $\frac{27}{32} = 0.84375$, no muss, no fuss, no Beta distributions or calculating the pdf of $X_{(2)}$ or simulations in R yielding three digits of accuracy with a million trials.

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    $\begingroup$ I’ve changed to an upvote. This is an interesting take on the problem, though I do think the purpose of this homework was to use order statistics. @Azka do you get how to work with the order statistic $X_{(2)}$? $\endgroup$
    – Dave
    Commented Sep 29, 2019 at 4:05
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    $\begingroup$ @Dave Yes I did work out through the order statistics method and got my answer. Thanks for your help. I understood the concept now. $\endgroup$
    – Azka
    Commented Sep 29, 2019 at 15:20
  • $\begingroup$ @Dilip Sarwate I think your method is quite interesting too just like Dave said. Its much easier on the computation part. Thanks for your help. $\endgroup$
    – Azka
    Commented Sep 29, 2019 at 15:24
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Comment: You have the math in Comments from @Dave and @whuber' link. The initial probability distribution is a beta distribution, specifically $\mathsf{Beta}(2,1).$ Maybe you can match your distribution for $H=X_{(2)}$ to results in the following simulation in R, using a million samples of size $n = 3.$ [With a million iterations you can expect 2, maybe 3, decimal places of accuracy.]

set.seed(928)
h = replicate(10^6, median(rbeta(3, 2, 1)))
mean(h > .5)
[1] 0.843916       # aprx P(H > .5)
2*sd(h>.5)/10^3
[1] 0.0007258703   # 95% margin of sim err for P(H > .5)
27/32
[1] 0.84375        # exact P(H > .5)   
mean(h)
[1] 0.685836
hist(h, prob=T, br=30, col="skyblue2")
abline(v = .5, lwd=2, col="red")

enter image description here

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