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11111nata11111 [884]
3 years ago
13

Suppose that Y has density function

Mathematics
1 answer:
zvonat [6]3 years ago
8 0

I'm assuming

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

(a) <em>f(x)</em> 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

<em>k</em> / 6 = 1   →   <em>k</em> = 6

(b) By definition of conditional probability,

P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = P(<em>Y</em> ≤ 0.4 and <em>Y</em> ≤ 0.8) / P(<em>Y</em> ≤ 0.8)

P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = P(<em>Y</em> ≤ 0.4) / P(<em>Y</em> ≤ 0.8)

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

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(<em>Y</em> ≤ 0.4) = <em>F</em> (0.4) = 0.352

P(<em>Y</em> ≤ 0.8) = <em>F</em> (0.8) = 0.896

and so

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

(c) The 0.95 quantile is the value <em>φ</em> such that

P(<em>Y</em> ≤ <em>φ</em>) = 0.95

In terms of the integral definition of the CDF, we have solve for <em>φ</em> 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

3<em>φ</em>² - 2<em>φ</em>³ = 0.95

Use a calculator to solve this and find that <em>φ</em> ≈ 0.865.

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