A continuous random variable X has the probability density f

A continuous random variable X has the probability density function (PDF) given by otherwise What is the numerical value of c? What is the expected value of A? What is the CDF of X?

Solution

A)

Normalizing,

Integral [4x dx]|(0, c) = 1

2x^2|(0, c) = 1

2c^2 = 1

c = sqrt(2)/2 or 0.707106781 [ANSWER]

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b)

E(x) = Integral [x f(x) dx]

E(x) = Integral [x(4x) dx]|(0, 0.707106781)

= 4 x^3 / 3 | (0, 0.707106781)

= 4(0.707106781)^3/3

E(x) = 0.47140452 [ANSWER]

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c)

F(x) = Integral [f(x) dx] + C, such that F(c) = 1.

Thus,

F(x) = 2x^2 + C

However,

F(sqrt(2)/2) = 2[sqrt(2)/2]^2 + C = 1

1 + C = 1

So

C = 0

Thus,

F(x) = 0, x<0
2x^2 0<=x<=0.47140452
1 x>0.47140452 [ANSWER]