Suppose that Y has density function
f(y) = ky(1 − y) if 0 ≤ y ≤ 1 0, elsewhere
A) Find the value of k that makes f (y) a probability density function.
B) Find P(Y ≤ .4|Y ≤ .8).
C) Find the .95-quantile, i.e., find a point φ.95 such that P(Y ≤ φ.95) = .95.

I'm assuming

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

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

[tex]\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[/tex]

Compute the integral:

[tex]\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[/tex]

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

[tex]\displaystyle F(y)=\int_{-\infty}^y f(t)\,\mathrm dt=\int_0^y6t(1-t)\,\mathrm dt=\begin{cases}0&\text{for }y<0\\3y^2-2y^3&\text{for }0\le y<1\\1&\text{for }y\ge1\end{cases}[/tex]

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

[tex]\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=0.95[/tex]

We have

[tex]\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[/tex]

which reduces to the cubic

3φ² - 2φ³ = 0.95

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

Using probability concepts, it is found that

a) The value is k = 6.

b) P(Y ≤ .4|Y ≤ .8) = 0.554.

c) The 95th percentile is x = 0.86465.

The density function is:

[tex]f(y) = ky(1 - y)[/tex]

Item a:

The condition that makes it a probability density function is:

[tex]\int_0^1 f(y) dy = 1[/tex]

Thus:

[tex]\int_0^1 f(y) dy = \int_0^1 ky - ky^2 dy[/tex]

[tex]\int_0^1 f(y) dy = k(\frac{y^2}{2} - \frac{y^3}{3})|_{y = 0}^{y = 1}[/tex]

[tex]\int_0^1 f(y) dy = k(\frac{1}{2} - \frac{1}{3})[/tex]

Then

[tex]k(\frac{1}{2} - \frac{1}{3}) = 1[/tex]

[tex]k(\frac{3-2}{6}) = 1[/tex]

[tex]k = 6[/tex]

The value is k = 6.

Item b:

This probability is:

[tex]\int_0.4^0.8 f(y) dy = 6(\frac{y^2}{2} - \frac{y^3}{3})|_{y = 0.4}^{y = 0.8}[/tex]

Then

[tex]p = 6\left(\frac{0.8^2}{2} - \frac{0.8^3}{3} - \frac{0.4^2}{2} + \frac{0.4^3}{3}\right)[/tex]

[tex]p = 0.554[/tex]

Thus, P(Y ≤ .4|Y ≤ .8) = 0.554.

Item c:

This is x for which:

[tex]\int_0^x f(y) dy = 0.95[/tex]

Thus:

[tex]6(\frac{x^2}{2} - \frac{x^3}{3}) = 0.95[/tex]

[tex]-2x^3 + 3x^2 - 0.95 = 0[/tex]

Solving a cubic equation with the help of a calculator, and considering [tex]0 \leq x \leq 1[/tex], this is x = 0.86465.

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