Question

In: Statistics and Probability

Let X be a continuous random variable with pdf: f(x) = ax^2 − 2ax, 0 ≤...

Let X be a continuous random variable with pdf: f(x) = ax^2 − 2ax, 0 ≤ x ≤ 2

(a) What should a be in order for this to be a legitimate p.d.f?

(b) What is the distribution function (c.d.f.) for X?

(d) What is the 90th percentile value of this distribution? (Note: If you do this problem correctly, you will end up with a cubic equation that you need to solve. You may solve it numerically.)

(e) What are the mean and variance of this distribution?

Solutions

Expert Solution

f(x) = ax² - 2ax ; 0 x 2

(a) Here for being a legitimate pdf

= 1

[ax³/3 - ax²]²₀ = 1

[8a/3 - 4a] = 1

a (8/3 - 4) = 1

a (-4/3) = 1

a = -3/4

(b) so here

f(x) = 3x/2 - 3x²/4

cdf for F(x) =

= [3x²/4 - 3x³/12] ^x₀

= 3x²/4 - x³/4

F(x) = -(x-3)x² /4

= (3x²/2 - 3x³/12)¹₀

= 0.5

P(x > 0.5) =

= (3x²/2 - 3x³/12)²_0.5

= 0.84375

Pr(x > 3) = 0

(d) Here let say 90 percentile of this distribution

0.9 = F(x₀)

0.9 = -(x-3)x² /4

3.6 = -x³ + 3x²

by solving x = 1.6084

(e) Here E[X] =

E{X] = 1

VaR[X} = E[X²] - E{X]²

E[X²] =

= 1.2

VaR[X[ = 1.2 - 1² = 0.2

orchestra answered 1 year ago

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