Question

Suppose that Y has density function

Answer 1

I'm assuming

$f(y)=\begin{cases}ky(1-y)&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(a) f(x) is a valid probability density function if its integral over the support is 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=k\int_0^1 y(1-y)\,\mathrm dy=k\int_0^1(y-y^2)\,\mathrm dy=1$

Compute the integral:

$\displaystyle\int_0^1(y-y^2)\,\mathrm dy=\left(\frac{y^2}2-\frac{y^3}3\right)\bigg|_0^1=\frac12-\frac13=\frac16$

So we have

k / 6 = 1   →   k = 6

(b) By definition of conditional probability,

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4 and Y ≤ 0.8) / P(Y ≤ 0.8)

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4) / P(Y ≤ 0.8)

It makes sense to derive the cumulative distribution function (CDF) for the rest of the problem, since F(y) = P(Y ≤ y).

We have

$\displaystyle F(y)=\int_{-\infty}^y f(t)\,\mathrm dt=\int_0^y6t(1-t)\,\mathrm dt=\begin{cases}0&\text{for }y$

Then

P(Y ≤ 0.4) = F (0.4) = 0.352

P(Y ≤ 0.8) = F (0.8) = 0.896

and so

P(Y ≤ 0.4 | Y ≤ 0.8) = 0.352 / 0.896 ≈ 0.393

(c) The 0.95 quantile is the value φ such that

P(Y ≤ φ) = 0.95

In terms of the integral definition of the CDF, we have solve for φ such that

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=0.95$

We have

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=\int_0^\varphi 6y(1-y)\,\mathrm dy=(3y^2-2y^3)\bigg|_0^\varphi = 0.95$

which reduces to the cubic

3φ² - 2φ³ = 0.95

Use a calculator to solve this and find that φ ≈ 0.865.