Verified solution · Exam P

SOA Exam P Sample Question #384 Solution

This Exam P sample reference tests Distribution Functions. This problem translates a percentile relation into the quantile function of a second distribution. The resulting CDF is x to the sixth power, whose derivative is 6x to the fifth power and choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Distribution Functions question

Setup

Normalize the first power density and integrate it on the unit interval.

f_1(x)=3x^2,\qquad 0\le x\le1

F_1(x)=x^3

Model

Invert the first CDF, then express the supplied percentile correspondence as a relation between quantile functions.

Q_1(p)=p^{1/3}

Q_2(p^2)=Q_1(p)=p^{1/3}

Compute

Replace the second percentile level by u, recover its quantile function, and invert it to obtain the CDF.

u=p^2\quad\Longrightarrow\quad Q_2(u)=u^{1/6}

F_2(x)=x^6

f_2(x)=F_2'(x)=6x^5

Answer

The second density is 6x to the fifth power on the unit interval.

\boxed{f_2(x)=6x^5\quad\text{(B)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis keeps the correct fifth-power shape but uses coefficient 2/3. Its integral over the unit interval is only 1/9, so it is not a density.
CThis adds the CDF powers three and two, then differentiates the resulting fifth-power CDF. Percentile composition multiplies the exponents instead.
DThis treats the second CDF as a square directly from the percentile label, ignoring the first distribution's cubic CDF.
EThis squares the first quantile instead of composing the percentile levels. Inversion then produces the wrong three-halves power density.

Original practice · fully worked

Original variant: density of a reported odds score

A device generates a raw calibration value U uniformly from 0 to 1. It reports the odds, U divided by one minus U, as score Y. Determine the density of Y for positive y.

A
y/(1+y)
B
1/(1+y)²
C
1/(1+y)
D
exp(-y)
E
1/y² for y>1

Variant answer in brief

The increasing odds transformation has inverse y divided by one plus y. This is also the transformed CDF, whose derivative is the inverse square of one plus y, choice B.

Setup

Solve the increasing reporting transformation for the raw value.

y=\frac{u}{1-u}

u=\frac{y}{1+y}

Model

Translate a reported-score event through the inverse mapping and use the uniform CDF.

F_Y(y)=\Pr\!\left(\frac{U}{1-U}\le y\right)

F_Y(y)=\Pr\!\left(U\le\frac{y}{1+y}\right)=\frac{y}{1+y},\qquad y>0

Compute

Differentiate the transformed CDF.

f_Y(y)=\frac{d}{dy}\left(\frac{y}{1+y}\right)=\frac1{(1+y)^2}

Answer

The odds score has an inverse-square density on the positive half-line.

\boxed{f_Y(y)=\frac1{(1+y)^2},\ y>0\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #384 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Distribution Functions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: density of a reported odds score

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution