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: Recognize the distribution without integrating for c first.
The form \( f(x) = ce^{-2x} \) for \( x > 0 \) is the standard exponential density \( f(x) = \lambda e^{-\lambda x} \) with rate \( \lambda = 2 \), so \( c \) must equal 2 for the total area to be 1.

Step 2: Use the known survival function of the exponential distribution.
For an exponential distribution with rate \( \lambda \), the probability of exceeding a value \( t \) has the standard closed form:
\[ P(X > t) = e^{-\lambda t} \]

Step 3: Substitute the given values directly.
With \( \lambda = 2 \) and \( t = 2 \):
\[ P(X > 2) = e^{-2(2)} = e^{-4} \]

Step 4: Final Answer.
\[ \boxed{P(X>2) = e^{-4}} \]
Therefore, option (C) is correct.