Suppose $X$ follows an exponential distribution with parameter $\lambda > 0$. Fix $a > 0$. Define the random variable $Y$ by $Y = k, \text{ if } ka \leq X < (k+1)a, \quad k = 0, 1, 2, \dots$ Which of the following statements are correct?
The problem involves a random variable X following an exponential distribution and defines another discrete random variable Y based on intervals of X. We need to determine the properties and distribution of Y.
Defining Random Variable Y
The random variable X follows an exponential distribution with PDF $f_X(x) = \lambda e^{-\lambda x}$ for $x \geq 0$. The variable Y is defined as:
- $Y = k$, if $ka \leq X < (k+1)a$, for $k = 0, 1, 2, \dots$
This definition implies that Y is a discrete random variable taking non-negative integer values ($0, 1, 2, \dots$).
Evaluating Option A: $P(4 < Y < 5) = 0$
Since Y can only take integer values ($0, 1, 2, \dots$), it is impossible for Y to fall strictly between 4 and 5.
Therefore, the probability $P(4 < Y < 5)$ must be 0.
Conclusion: Option A is correct.
Evaluating Option B: Y follows an exponential distribution
An exponential distribution is a continuous probability distribution.
As established, the random variable Y is discrete.
A discrete random variable cannot follow a continuous distribution.
Conclusion: Option B is incorrect.
Evaluating Option C: Y follows a geometric distribution
We calculate the probability mass function (PMF) for Y.
For any integer $k \geq 0$:
$ P(Y=k) = P(ka \leq X < (k+1)a) $
$ P(Y=k) = \int_{ka}^{(k+1)a} \lambda e^{-\lambda x} dx $
$ P(Y=k) = \left[ -e^{-\lambda x} \right]_{ka}^{(k+1)a} $
$ P(Y=k) = -e^{-\lambda (k+1)a} - (-e^{-\lambda ka}) $
$ P(Y=k) = e^{-\lambda ka} - e^{-\lambda (k+1)a} $
Factor out $e^{-\lambda ka}$:
$ P(Y=k) = e^{-\lambda ka} (1 - e^{-\lambda a}) $
Let $p = 1 - e^{-\lambda a}$. Since $\lambda > 0$ and $a > 0$, we have $0 < e^{-\lambda a} < 1$, which implies $0 < p < 1$.
Also, let $q = e^{-\lambda a} = 1 - p$. Note that $0 < q < 1$.
The PMF becomes:
$ P(Y=k) = q^k \cdot p $
$ P(Y=k) = p \cdot (1-p)^k $
This is the probability mass function of a geometric distribution defined for $k = 0, 1, 2, \dots$, with parameter $p = 1 - e^{-\lambda a}$.
Conclusion: Option C is correct.
Evaluating Option D: Y follows a Poisson distribution
The PMF of a Poisson distribution is given by $P(Y=k) = \frac{e^{-\mu} \mu^k}{k!}$ for $k = 0, 1, 2, \dots$.
The derived PMF for Y is $P(Y=k) = (1 - e^{-\lambda a}) (e^{-\lambda a})^k$.
This form does not match the Poisson PMF (it lacks the $k!$ term and the structure is different).
Conclusion: Option D is incorrect.
Final Result
Based on the analysis, the correct statements are A and C.
Important Questions from Random Variables
Let $X$, $Y$, and $Z$ be independent Normal random variables with means $-1$, $0$, and $1$, respectively, and variances $1$, $1$, and $3$, respectively. Which of the following random variables has a Cauchy distribution with location parameter $0$ and scale parameter $1$?
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Let $X_1$ and $X_2$ be a random sample from Uniform$[0, \theta]$ distribution, where $\theta > 0$. For testing the hypothesis
$H_0: \theta = 1$ against $H_1: \theta = 2$,
consider a test which rejects $H_0$ if $X_1 + X_2 > \frac{4}{5}$. Then, the probability of type-I error is- Let $\{Y_n: n \ge 1\}$ be a sequence of independent and identically distributed random variables, where $Y_1 \sim \text{Bernoulli}(\frac{1}{2})$. Define $Z = \sum_{n=1}^\infty \frac{4Y_n}{5^n}$. Then, which of the following statements is true?
Let $X$ be a single sample from an absolutely continuous distribution with probability density function
$f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise,} \end{cases}$
where $\theta > 0$ is unknown. Which of the following intervals is a $95\%$ confidence interval for $\theta$?
