Let [math]X[/math] be a random variable with density function

[[math]] f(x) = \begin{cases} 2e^{-2x}, \, x \gt0 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate [math]\operatorname{P}[ X \leq 0.5 | X \leq 1.0].[/math]

• 0.433
• 0.547
• 0.632
• 0.731
• 0.865

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Understand the Problem and Conditional Probability

The problem asks us to calculate the conditional probability [math]\operatorname{P}[ X \leq 0.5 | X \leq 1.0][/math] for a random variable [math]X[/math] with a given density function. The density function is:

[[math]] f(x) = \begin{cases} 2e^{-2x}, \, x \gt0 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Conditional Probability Formula: The formula for conditional probability is given by:

[[math]]\operatorname{P}[A|B] = \frac{\operatorname{P}[A \cap B]}{\operatorname{P}[B]}[[/math]]

Identify Events A and B: In this problem, we define our events as:

• [math]A = \{X \leq 0.5\}[/math]
• [math]B = \{X \leq 1.0\}[/math]

Determine the Intersection of Events: Since [math]0.5 \leq 1.0[/math], the event [math]A \cap B[/math] (where [math]X \leq 0.5[/math] AND [math]X \leq 1.0[/math]) simplifies to [math]X \leq 0.5[/math]. Therefore, [math]A \cap B = \{X \leq 0.5\}[/math]. Rewrite the Conditional Probability: Substituting this into the conditional probability formula, we get:

[[math]]\operatorname{P}[ X \leq 0.5 | X \leq 1.0] = \frac{\operatorname{P}[X \leq 0.5]}{\operatorname{P}[X \leq 1.0]}[[/math]]

Relate to the Cumulative Distribution Function (CDF): The probability [math]\operatorname{P}[X \leq a][/math] is by definition the Cumulative Distribution Function (CDF) evaluated at [math]a[/math], denoted as [math]F(a)[/math]. So, the expression becomes:

[[math]]\operatorname{P}[ X \leq 0.5 | X \leq 1.0] = \frac{F(0.5)}{F(1.0)}[[/math]]

Step 2: Calculate the Cumulative Distribution Function (CDF)

The CDF, [math]F(x)[/math], is the integral of the probability density function (PDF), [math]f(x)[/math], from [math]-\infty[/math] to [math]x[/math]. For [math]x \gt 0[/math], given that [math]f(x) = 2e^{-2x}[/math] for [math]x \gt 0[/math] and [math]0[/math] otherwise, we calculate [math]F(x)[/math] as follows:

[[math]] F(x) = \int_0^{x} 2e^{-2t} dt [[/math]]

Perform the Integration: To integrate [math]2e^{-2t}[/math], we use a simple substitution or direct integration:

[[math]] \int 2e^{-2t} dt = 2 \left( -\frac{1}{2} e^{-2t} \right) = -e^{-2t} [[/math]]

Now, evaluate the definite integral from [math]0[/math] to [math]x[/math]:

[[math]] F(x) = \left[ -e^{-2t} \right]_0^x = (-e^{-2x}) - (-e^{-2 \cdot 0}) = -e^{-2x} - (-e^0) = -e^{-2x} - (-1) = 1 - e^{-2x} [[/math]]

Thus, the CDF for [math]x \gt 0[/math] is:

[[math]] F(x) = 1 - e^{-2x} [[/math]]

Step 3: Evaluate the CDF at the Required Points

Now we need to evaluate [math]F(x)[/math] at [math]x = 0.5[/math] and [math]x = 1.0[/math]. Calculate [math]F(0.5)[/math]: Substitute [math]x = 0.5[/math] into the CDF formula:

[[math]] F(0.5) = 1 - e^{-2(0.5)} = 1 - e^{-1} [[/math]]

Calculate [math]F(1.0)[/math]: Substitute [math]x = 1.0[/math] into the CDF formula:

[[math]] F(1.0) = 1 - e^{-2(1.0)} = 1 - e^{-2} [[/math]]

Step 4: Calculate the Conditional Probability

Now we can substitute the evaluated CDF values back into the conditional probability expression derived in Step 1:

[[math]] \operatorname{P}[ X \leq 0.5 | X \leq 1.0] = \frac{F(0.5)}{F(1.0)} = \frac{1 - e^{-1}}{1 - e^{-2}} [[/math]]

Numerical Calculation: To get the numerical result, we approximate the values of [math]e^{-1}[/math] and [math]e^{-2}[/math]:

• [math]e^{-1} \approx 0.367879[/math]
• [math]e^{-2} \approx 0.135335[/math]

Substitute these values:

[[math]] \operatorname{P}[ X \leq 0.5 | X \leq 1.0] = \frac{1 - 0.367879}{1 - 0.135335} = \frac{0.632121}{0.864665} \approx 0.731089 [[/math]]

Rounding to three decimal places, the result is [math]0.731[/math].

Key Insights

• The calculation of conditional probabilities for continuous random variables relies on understanding the intersection of events and the Cumulative Distribution Function (CDF).
• When dealing with [math]\operatorname{P}[X \leq a | X \leq b][/math] where [math]a \leq b[/math], the intersection simplifies to [math]X \leq a[/math], allowing the formula to be expressed as [math]\frac{F(a)}{F(b)}[/math].
• The CDF is found by integrating the PDF. For an exponential distribution starting at 0, the CDF is typically of the form [math]1 - e^{-\lambda x}[/math].
• Accurate calculation of the definite integral is crucial for obtaining the correct CDF.
• The problem implicitly uses properties of the exponential distribution, a common distribution in probability and statistics.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: D

[[math]] \begin{align*} F(x) &= \int_0^{x}2e^{-2y} dy = -e^{-2y} \Big |_0^x = 1-e^{-2x} \\ \operatorname{P}[X \leq 0.5 | X \leq 1.0] &= \frac{\operatorname{P}[X \leq 0.5]}{\operatorname{P}[X \leq 1.0]} = \frac{F(0.5)}{F(1.0)} = \frac{1-e^{-1}}{1-e^{-2}} = 0.731. \end{align*} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.