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If the pdf for a continuous random variable (x) = 2x^(−3) for 1

< x < [infinity], find the
a. Median
b. E[X]
c. V[X]
d. P(>5|>4)

1 Answer

4 votes

Final answer:

To find the median, E[X], and V[X] for the given pdf of a continuous random variable, the appropriate integrations over the specified intervals must be calculated. The conditional probability P(>5|>4) requires dividing the probability that x is greater than 5 by the probability that x is greater than 4.

Step-by-step explanation:

The continuous random variable x has a probability density function (pdf): 2x^(−3) for the range 1 < x < infinity. To answer the questions, we need to perform calculations with this pdf.

a. Median: The median m is the value such that the probability of x being less than m is 0.5. That is, we need to solve the equation Integrate[2x^(−3), {x, 1, m}] = 0.5.

b. E[X]: The expected value E[X] is calculated as Integrate[x * 2x^(−3), {x, 1, infinity}]. However, this integral diverges, indicating that the mean does not exist for this distribution.

c. V[X]: The variance V[X] is calculated as E[X^2] - (E[X])^2. But since E[X] does not exist, neither does V[X].

d. P(>5|>4): The conditional probability of x being greater than 5 given that it is greater than 4 is calculated using the formula P(A|B) = P(A and B) / P(B). For continuous distributions, P(x > 5) is found by integrating the pdf from 5 to infinity, and P(x > 4) is found by integrating from 4 to infinity.

We can not provide the exact numerical values without the proper tools to perform the integration, but the methodology outlined would yield the answer given appropriate computation.

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