ISS 2016 Statistics Paper-1 Solution: Question 5 (Harmonic Mean of Beta Type II Distribution)
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Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, we now arrive at Question 5 of the 2016 paper. This one drops a slightly unusual distribution on you — the Beta distribution of the second kind — and asks for a summary measure that most students only ever practise on discrete data: the harmonic mean.
Quick Summary
Topic: Harmonic mean of a continuous distribution (Beta distribution of the second kind / Beta prime distribution)
Question reference: ISS 2016, Statistics Paper-1, Question 5
Correct answer: Option (c), (p − 1)/q
The Question, In Full
For the distribution
f(x) = [1/β(p,q)] · xp−1/(1+x)p+q, 0 < x < ∞, p > 0, q > 0; what is the harmonic mean?
(a) p/(p+q)
(b) 1/p
(c) (p − 1)/q
(d) (p + 1)/(q − 1)
Step 1: Recognise the Distribution
The density f(x) = xp−1(1+x)−(p+q)/β(p,q), for x > 0, is the standard form of the Beta distribution of the second kind (also called the Beta prime distribution). It is closely related to the F-distribution, which is why it keeps turning up in this part of the syllabus. The constant 1/β(p,q) is just there to make the total area under the curve equal to 1, using the Beta function β(p,q) = Γ(p)Γ(q)/Γ(p+q).
Step 2: Recall What "Harmonic Mean" Means for a Continuous Variable
For a set of numbers, the harmonic mean is the reciprocal of the average of the reciprocals. For a continuous random variable X, the exact same idea applies, except "average of the reciprocals" becomes an expectation:
H = 1 / E(1/X) = 1 / E(X−1)
So the entire problem boils down to computing E(X−1) for this distribution, and then flipping it over.
Step 3: Set Up the Integral for E(X−1)
By definition,
E(X−1) = ∫0∞ x−1 · f(x) dx = [1/β(p,q)] ∫0∞ xp−2 (1+x)−(p+q) dx
(We simply combined x−1 with xp−1 from the density to get xp−2.)
Step 4: Use the Standard Beta Integral
This is the one fact you need to memorise to solve almost any moment question on this distribution:
∫0∞ xa−1 (1+x)−(a+b) dx = β(a, b), for a > 0, b > 0.
Look at our integral: ∫0∞ xp−2 (1+x)−(p+q) dx. Matching powers, we need xa−1 = xp−2, so a − 1 = p − 2, which gives a = p − 1. And we need a + b = p + q, so b = (p+q) − a = (p+q) − (p−1) = q + 1.
So the integral equals β(p−1, q+1), and therefore:
E(X−1) = β(p−1, q+1) / β(p,q)
Pro Tip: The trickiest part of this question is not the calculus — it's correctly matching exponents to find the new Beta parameters. Write a − 1 = (power of x) and a + b = (power of (1+x), with sign flipped) as two separate equations every single time, rather than trying to "read off" the shifted parameters directly. Students who drill this exponent-matching step until it becomes automatic rarely lose marks on Beta-type-II or F-distribution moment questions, because the same two-equation trick solves E(X), E(X²), and E(1/X) alike — only the target power of x changes.
Step 5: Expand the Beta Functions Using Gamma Functions
Recall β(a,b) = Γ(a)Γ(b)/Γ(a+b). Since both our Beta functions have a+b = p+q, that part cancels neatly:
E(X−1) = [Γ(p−1)Γ(q+1)/Γ(p+q)] ÷ [Γ(p)Γ(q)/Γ(p+q)] = [Γ(p−1)Γ(q+1)] / [Γ(p)Γ(q)]
Now use the basic Gamma function recurrence Γ(n+1) = n·Γ(n), applied twice:
Since Γ(p) = (p−1)·Γ(p−1), we get Γ(p−1)/Γ(p) = 1/(p−1)
Since Γ(q+1) = q·Γ(q), we get Γ(q+1)/Γ(q) = q
Multiplying these two results together:
E(X−1) = q / (p − 1)
Step 6: Take the Reciprocal to Get the Harmonic Mean
Since H = 1/E(X−1):
H = (p − 1) / q
This matches option (c)exactly.
Step 7: Confirm Against the Official Answer Key
The official ISS 2016 Statistics Paper-1 answer key lists the correct response to Question 5 as (c), which agrees precisely with the independent derivation above.The harmonic mean of this Beta distribution of the second kind is (p − 1)/q, and both the calculation and the key are in full agreement — there is no discrepancy to flag here.
Why This Question Matters
This question is a good test of whether a candidate actually understands moments as integrals, rather than having memorised a formula sheet. The Beta distribution of the second kind rarely gets its own dedicated chapter in introductory textbooks, yet it underlies the F-distribution used constantly in ANOVA and regression theory. Being able to derive E(Xk) for any integer k — positive or negative — from the single Beta-integral identity is a transferable skill: the same method finds the mean, variance, and even the mode of this family, not just the harmonic mean asked here.
Frequently Asked Questions
What exactly is the Beta distribution of the second kind?
It is a continuous distribution on (0, ∞) with density proportional to xp−1/(1+x)p+q, governed by two positive shape parameters p and q. It is also called the Beta prime distribution, and if X follows it, then X/(1+X) follows an ordinary Beta(p, q) distribution on (0, 1).
Why can't we just find E(X) first and then take its reciprocal?
Because the harmonic mean is defined as the reciprocal of the average of reciprocals, not the reciprocal of the average. In symbols, 1/E(X) is generally not equal to E(1/X), so you must compute E(1/X) directly and then invert it — taking a shortcut here gives a wrong answer that doesn't match any of the options.
Does the harmonic mean of this distribution always exist?
No — the derivation requires p − 1 > 0, i.e. p > 1, for the integral defining E(X−1) to converge. If p ≤ 1, the harmonic mean is not finite. The question implicitly assumes p is large enough for the expression to make sense.
Is the Beta distribution of the second kind related to the F-distribution?
Yes, very closely. If X follows a Beta Type II distribution with parameters p = m/2 and q = n/2, then (n/m)X follows an F-distribution with (m, n) degrees of freedom. That is one reason moment identities for this family show up so often in inferential statistics questions.
What is the single most useful formula to remember for this whole family of questions?
The identity ∫0∞ xa−1(1+x)−(a+b) dx = β(a,b). Once you can match exponents to find a and b for any moment E(Xk), the rest of the problem is just Gamma function bookkeeping.
What is a common mistake candidates make on questions like this?
The most frequent slip is misreading E(X−1) as requiring a − 1 = p and mismatching the second parameter, which throws off the final answer by a shifted index. Carefully writing out both matching equations — for the power of x and for the combined power of (1+x) — avoids this entirely.
If any step above still feels shaky, don't let it slide — drop your doubt in the comments and we'll work through it together, and if you know a fellow ISS aspirant who's stuck on this same question, do pass this along to them.
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