Question

If the density function of X equals f(x) \(=\) \(\begin{cases} ce^{-2x}, & 0 \lt x \lt \infty \\ 0, & x \lt 0 \end{cases}\), then P(X \(\gt \) 2) is

• A. 1 \(-\) e\(^{-4}\)
• B. 1 \(-\) e\(^{-2}\)
• C. e\(^{-4}\)
• D. e\(^{-2}\)

Hint:

For any standard exponential distribution with PDF \( f(x) = \lambda e^{-\lambda x} \) for \( x \gt 0 \):
- The cumulative distribution function (CDF) is \( P(X \le x) = 1 - e^{-\lambda x} \).
- The survival function is \( P(X \gt x) = e^{-\lambda x} \).
Using this formula directly: \( P(X \gt 2) = e^{-2(2)} = e^{-4} \). This avoids integration during exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks to calculate a probability \( P(X \gt 2) \) for a continuous random variable \( X \) defined by an exponential-type probability density function (PDF).

Step 2: Key Formula or Approach:
For a PDF \( f(x) \) to be valid, the total area under the curve must equal 1:
\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \]
Once the normalization constant \( c \) is found, the probability of an event \( X \gt 2 \) is calculated by integrating the PDF:
\[ P(X \gt 2) = \int_2^{\infty} f(x) \, dx \]

Step 3: Detailed Explanation:

• Find the Normalization Constant \( c \):
\[ \int_0^{\infty} c e^{-2x} \, dx = 1 \] \[ c \left[ \frac{e^{-2x}}{-2} \right]_0^{\infty} = 1 \] \[ -\frac{c}{2} \left[ e^{-\infty} - e^0 \right] = 1 \implies -\frac{c}{2} [0 - 1] = 1 \] \[ \frac{c}{2} = 1 \implies c = 2 \] - Thus, \( f(x) = 2e^{-2x} \) for \( x \gt 0 \), which is a standard exponential distribution with parameter \( \lambda = 2 \).

• Calculate \( P(X \gt 2) \):
\[ P(X \gt 2) = \int_2^{\infty} 2 e^{-2x} \, dx \] \[ P(X \gt 2) = 2 \left[ \frac{e^{-2x}}{-2} \right]_2^{\infty} \] \[ P(X \gt 2) = -1 \left[ e^{-\infty} - e^{-4} \right] \] \[ P(X \gt 2) = -1 \left[ 0 - e^{-4} \right] = e^{-4} \]

Step 4: Final Answer:
The probability \( P(X \gt 2) \) is exactly \( e^{-4} \).
Therefore, the correct choice is option (C).