Let X be a random variable that takes non-zero values in [1,∞), with a PDF of the form
fX(x)=⎧⎩⎨cx3,0,if x≥1,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=
- unanswered
P(X≤U)=
- unanswered
Find the PDF of D=1/X. Express your answer in terms of d using standard notation .
For 0≤d≤1, fD(d)=
To find the value of the constant c, we need to use the fact that the probability density function (PDF) integrates to 1 over its support. In this case, the support is [1,∞), so we have:
∫[1,∞) cx^3 dx = 1
To integrate cx^3, we increase the power by 1 and divide by the new power:
c/4 * x^4 |[1,∞) = 1
Substituting the limits of integration, we get:
c/4 * (∞^4 - 1^4) = 1
Since ∞^4 = ∞ and c/4 * (∞ - 1) = 1, we can solve for c:
c/4 * ∞ - c/4 = 1
As ∞ is not a well-defined value, we can ignore it in this case. Thus, we have:
-c/4 = 1
Solving for c, we get:
c = -4
Now let's move on to the next question.
To find P(X ≤ U), we need to calculate the cumulative distribution function (CDF) of X and evaluate it at the value of U. The CDF, F(x), gives the probability that X takes a value less than or equal to x.
First, let's find the CDF of X. Since X is non-zero in [1,∞), we can calculate it as:
F(x) = ∫[1,x] cx^3 dx
Integrating cx^3, we get:
c/4 * x^4 |[1,x] = c/4 * (x^4 - 1^4)
Evaluating the CDF at U, we have:
P(X ≤ U) = F(U) = c/4 * (U^4 - 1)
Now, let's move on to the last question.
To find the probability density function (PDF) of D = 1/X, we need to use the transformation method. The general formula for transforming a random variable is:
fD(d) = fX(g(d)) * |(dg(d)/dd)|
In this case, we have D = 1/X, so g(d) = 1/d. We also need to find |(dg(d)/dd)|, which is the absolute value of the derivative of g(d) with respect to d:
|(dg(d)/dd)| = |d(-1/d^2)/dd| = |-1/d^2| = 1/d^2
Substituting these values into the formula, we have:
fD(d) = fX(1/d) * 1/d^2
Since the PDF of X is given by fX(x) = cx^3, we can substitute 1/d for x:
fD(d) = c(1/d)^3 * 1/d^2 = c/d^5
Therefore, for 0 ≤ d ≤ 1, the PDF of D is fD(d) = c/d^5.