Math Problem Statement

Let X be exponential with mean 1/λ; that is, fX (x)= λe^(-λx),0<x<∞ Find Ε[X|X>1].

Solution

To find E[XX>1]\mathbb{E}[X \mid X > 1], we need to compute the conditional expectation of XX given that X>1X > 1. The random variable XX follows an exponential distribution with mean 1/λ1/\lambda, which has the probability density function (PDF) given by:

fX(x)=λeλx,0<x<f_X(x) = \lambda e^{-\lambda x}, \quad 0 < x < \infty

The conditional expectation E[XX>1]\mathbb{E}[X \mid X > 1] is defined as:

E[XX>1]=1xfX(x)dxP(X>1)\mathbb{E}[X \mid X > 1] = \frac{\int_1^{\infty} x f_X(x) \, dx}{P(X > 1)}

Step 1: Calculate P(X>1)P(X > 1)

The probability P(X>1)P(X > 1) is given by the survival function of the exponential distribution:

P(X>1)=1λeλxdxP(X > 1) = \int_1^{\infty} \lambda e^{-\lambda x} \, dx

This integral can be computed as:

P(X>1)=[eλx]1=eλP(X > 1) = \left[ -e^{-\lambda x} \right]_1^{\infty} = e^{-\lambda}

Step 2: Calculate the numerator 1xfX(x)dx\int_1^{\infty} x f_X(x) \, dx

The next step is to calculate the integral:

1xλeλxdx\int_1^{\infty} x \lambda e^{-\lambda x} \, dx

This integral can be solved using integration by parts. Let u=xu = x and dv=λeλxdxdv = \lambda e^{-\lambda x} dx, then:

1xλeλxdx=[xeλx]1+1eλxdx\int_1^{\infty} x \lambda e^{-\lambda x} \, dx = \left[ -x e^{-\lambda x} \right]_1^{\infty} + \int_1^{\infty} e^{-\lambda x} \, dx

Evaluating the boundary terms and the remaining integral:

[xeλx]1=0(1eλ)=eλ\left[ -x e^{-\lambda x} \right]_1^{\infty} = 0 - (-1 e^{-\lambda}) = e^{-\lambda} 1eλxdx=eλλ\int_1^{\infty} e^{-\lambda x} \, dx = \frac{e^{-\lambda}}{\lambda}

Thus, the total integral is:

1xλeλxdx=eλ+eλλ\int_1^{\infty} x \lambda e^{-\lambda x} \, dx = e^{-\lambda} + \frac{e^{-\lambda}}{\lambda}

Step 3: Calculate E[XX>1]\mathbb{E}[X \mid X > 1]

Now, using the formula for conditional expectation:

E[XX>1]=eλ+eλλeλ=1+1λ\mathbb{E}[X \mid X > 1] = \frac{e^{-\lambda} + \frac{e^{-\lambda}}{\lambda}}{e^{-\lambda}} = 1 + \frac{1}{\lambda}

Thus, the conditional expectation is:

E[XX>1]=1+1λ\mathbb{E}[X \mid X > 1] = 1 + \frac{1}{\lambda}

Final Answer:

E[XX>1]=1+1λ\mathbb{E}[X \mid X > 1] = 1 + \frac{1}{\lambda}

Do you have any questions or would you like further details on any of the steps?

Here are 5 related questions:

  1. What is the general formula for the conditional expectation of an exponential random variable given a threshold X>aX > a?
  2. How is the survival function of an exponential distribution derived?
  3. Can you explain the intuition behind the exponential distribution's memoryless property?
  4. How does the mean of an exponential distribution change if the rate parameter λ\lambda is doubled?
  5. What is the variance of the conditional distribution of XX given X>1X > 1?

Tip: When solving problems involving exponential distributions, remember that the exponential distribution is memoryless, meaning past events do not influence future outcomes.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Exponential Distribution
Conditional Expectation

Formulas

f_X(x) = λe^(-λx), 0 < x < ∞ (PDF of exponential distribution)
P(X > 1) = e^(-λ)
Ε[X|X > 1] = 1 + 1/λ

Theorems

Conditional Expectation Formula
Survival Function of Exponential Distribution
Integration by Parts

Suitable Grade Level

Undergraduate