Suppose that X has a PDF of the form
f(x) {1/x^2 if x>= 1
{ 0 otherwise
For any x>2, the conditional PDF of X, given the event X>2 is
(Your answer should be an algebraic function of x)
2/x^2
oki
Is there anything else you would like to know or ask?
Oh, conditional probability, how I love to calculate you. Let's dive into this question, shall we?
So, we have this PDF of X given as f(x) = 1/x^2 for x ≥ 1 and 0 otherwise. We want to find the conditional PDF of X, given the event X > 2.
To find the conditional PDF, we need to consider the restrictions imposed by the event X > 2. Any value of X less than or equal to 2 would not satisfy the condition X > 2, so our conditional PDF will be 0 for x ≤ 2.
For x > 2, we need to normalize the PDF so that its integral over the range x > 2 is equal to 1 (since it represents a valid probability distribution).
The normalization constant can be found by integrating the original PDF over the range x > 2. So, let's do some math wizardry.
∫(1/x^2)dx from 2 to infinity = -1/x evaluated from 2 to infinity
= (-1/infinity) - (-1/2)
= 0 - (-1/2)
= 1/2
Now, to get the conditional PDF, we divide the original PDF by the total probability (our normalization constant):
g(x) = (1/x^2) / (1/2) = 2/x^2
So, the conditional PDF of X, given the event X > 2, is g(x) = 2/x^2 for x > 2 and 0 otherwise.
Voila! We have our algebraic function for the conditional PDF. Enjoy the mathematical circus!
To find the conditional PDF of X given the event X > 2, we need to calculate the conditional probability density function, denoted as f(x | X > 2).
The conditional PDF is defined as the ratio of the joint PDF of X and the event X > 2, divided by the probability of the event X > 2. It can be expressed as:
f(x | X > 2) = f(x, X > 2) / P(X > 2)
First, let's calculate the joint PDF of X and the event X > 2, denoted as f(x, X > 2). Since X has a PDF of 1/x^2 for x >= 1 and 0 otherwise, we need to consider the range in which X > 2 holds.
If X > 2, it means X must be greater than or equal to 2. Therefore, we have:
f(x, X > 2) = f(x) for x >= 2
Since the joint PDF is only non-zero when X >= 2 in this case.
Next, let's calculate the probability of the event X > 2, denoted as P(X > 2). This can be found by integrating the PDF of X over the range X > 2:
P(X > 2) = ∫(2 to ∞) f(x) dx
Using the given PDF of f(x) = 1/x^2 for x >= 1, we need to integrate it from 2 to infinity.
After finding P(X > 2), we can substitute it back into the conditional PDF equation to get the final answer, which will be an algebraic function of x.
Keep in mind that without knowing the exact range of integration for the given problem, we cannot provide a specific algebraic function as the answer.