Let X be a random variable that takes non-zero values in [1,∞), with a PDF of the form

fX(x)=⎧⎩⎨cx3 if x≥1, 0,otherwise.

Let U be a uniform random variable on [0,2]. Assume that X and U are independent.

What is the value of the constant c?

c=

P(X≤U)=

Find the PDF of D=1/X. Express your answer in terms of d using standard notation.

For 0≤d≤1, fD(d)=

5 answers

The line was supposed to be

fX(x)=c/x^3 if x≥1, 0,otherwise.

c = 2

1 c=2

2 ???
3 d*2

2. P(X≤U)= 0.25

To find the value of the constant c, we can use the fact that the total area under a probability density function (PDF) must equal 1.

Since X takes non-zero values in [1,∞), we can find the constant c by integrating the PDF over its entire range and setting it equal to 1:

1 = ∫[1,∞) c*x^3 dx

Integrating c*x^3 with respect to x gives:

1 = [c*(x^4)/4] evaluated from 1 to ∞

Since the upper limit is ∞, we can evaluate the limit as x approaches ∞:

1 = [c*(∞^4)/4] - [c*(1^4)/4]

Since (∞^4) is not a determinable value, we can say that it approaches infinity. So, we can ignore this term:

1 = - [c/4]

Now, we can solve for c:

c = -4

Next, to find P(X ≤ U), we need to compare the cumulative distribution function (CDF) of X and the CDF of U.

The CDF of X is found by integrating the PDF:

FX(x) = ∫[1,x] -4*t^3 dt

FX(x) = -t^4 evaluated from 1 to x

FX(x) = -(x^4) + 1

The CDF of U is a uniform distribution, which is a straight line from 0 to 2:

FU(u) = u/2

To find P(X ≤ U), we substitute U into the CDF of X:

P(X ≤ U) = P(-(U^4) + 1) = U/2

So, P(X ≤ U) = U/2.

Lastly, we need to find the PDF of the random variable D = 1/X. We can use the method of transformation to find the PDF of D.

To do this, we need to find the CDF of D, FD(d), and then differentiate it to obtain the PDF.

The CDF of D can be expressed as:

FD(d) = P(D ≤ d) = P(1/X ≤ d) = P(X ≥ 1/d)

Since X has a lower bound of 1, we can rewrite this as:

FD(d) = 1 - P(X < 1/d) = 1 - FX(1/d)

Substituting the CDF of X we found earlier:

FD(d) = 1 - (-(1/d)^4 + 1) = 1 + (1/d^4) - 1 = 1/d^4

Now, we differentiate FD(d) with respect to d to obtain the PDF of D:

fD(d) = dF'(d)/dd = d/d [1/d^4] = -4/d^5

Therefore, for 0 ≤ d ≤ 1, the PDF of D is fD(d) = -4/d^5.