Question

Let \( X \) be a random variable with probability density function \[ f(x) = \begin{cases} \frac{1}{x^2} & \text{if } x \geq 1, \\ 0 & \text{otherwise.} \end{cases} \] If \( Y = \log X \), then \( P(Y < 1 \mid Y < 2) \) equals

• A. $\frac{e}{1 + e}$
• B. $\frac{e - 1}{e + 1}$
• C. $\frac{1}{1 + e}$
• D. $\frac{1}{e - 1}$

Hint:

- The density of a transformed random variable is computed using the Jacobian of the transformation.
- For a logarithmic transformation, the new density function follows the rule: $f_Y(y) = f_X(e^y) . e^y$.

Answer 1

The Correct Option is A

1) Understanding the relationship between $X$ and $Y$:
We are given that $Y = \log X$. This means that $X = e^Y$. The probability density function of $X$ is: \[ f_X(x) = \frac{1}{x^2}, x \geq 1. \] Thus, for $Y = \log X$, the probability density function of $Y$ becomes: \[ f_Y(y) = f_X(e^y) . \frac{d}{dy} e^y = \frac{1}{e^{2y}} . e^y = \frac{1}{e^y}, y \geq 0. \] 2) Computing $P(Y<1 \mid Y<2)$:
The conditional probability $P(Y<1 \mid Y<2)$ is given by: \[ P(Y<1 \mid Y<2) = \frac{P(Y<1 \cap Y<2)}{P(Y<2)} = \frac{P(Y<1)}{P(Y<2)}. \] Since $Y$ has the probability density function $f_Y(y) = \frac{1}{e^y}$ for $y \geq 0$, we can compute the probabilities as follows:
- \( P(Y<1) = \int_0^1 \frac{1}{e^y} \, dy = 1 - \frac{1}{e}. \)
- \( P(Y<2) = \int_0^2 \frac{1}{e^y} \, dy = 1 - \frac{1}{e^2}. \)
Thus, \[ P(Y<1 \mid Y<2) = \frac{1 - \frac{1}{e}}{1 - \frac{1}{e^2}} = \frac{e - 1}{e + 1}. \] 3) Final Answer:
The correct answer is (A) $\frac{e}{1 + e}$.