Question

For \( x \in \mathbb{R} \), the floor function is denoted by \( f(x) = \lfloor x \rfloor \) and defined as follows \[ \lfloor x \rfloor = k, \quad k \leq x<k + 1, \] where \( k \) is an integer. Let \( Y = |X| \), where \( X \) is an exponentially distributed random variable with mean \( \frac{1}{\ln 10} \), where \( \ln \) denotes natural logarithm. For any positive integer \( \ell \), one can write the probability of the event \( Y = \ell \) as follows: \[ P(Y = \ell) = q^\ell (1 - q) \] The value of \( q \) is:

• A. 0.1
• B. 0.01
• C. 0.5
• D. 0.434

Hint:

For exponential distributions, the probability of a value lying within a specific range can be computed using the CDF. The formula for the floor function can be used to compute discrete probabilities for the transformed random variable.

Answer 1

The Correct Option is A

We are given that \( X \) follows an exponential distribution with mean \( \frac{1}{\ln 10} \), and \( Y = |X| \). The probability \( P(Y = \ell) \) can be derived using the CDF of the exponential distribution: \[ P(Y = \ell) = P(\ell \leq X<\ell + 1) = F_X(\ell + 1) - F_X(\ell) \] The CDF of \( X \) is: \[ F_X(x) = 1 - e^{-x \ln 10} \] So, the probability is: \[ P(Y = \ell) = e^{-\ell \ln 10} \left( 1 - e^{-\ln 10} \right) \] Comparing this with \( P(Y = \ell) = q^\ell (1 - q) \), we find: \[ q = e^{-\ln 10} = \frac{1}{10} \] Thus, the value of \( q \) is \( \boxed{0.1} \).