Understanding Exponential Random Variables
We are given two independent exponential random variables, $X$ and $Y$. Their expected values are $E[X] = 1$ and $E[Y] = \frac{1}{2}$.
- For an exponential distribution, the rate parameter $\lambda$ is the reciprocal of the expected value.
- Rate parameter for $X$: $\lambda_X = \frac{1}{E[X]} = \frac{1}{1} = 1$. The probability density function (PDF) is $f_X(x) = e^{-x}$ for $x \ge 0$.
- Rate parameter for $Y$: $\lambda_Y = \frac{1}{E[Y]} = \frac{1}{1/2} = 2$. The PDF is $f_Y(y) = 2e^{-2y}$ for $y \ge 0$.
- Since $X$ and $Y$ are independent, their joint PDF is $f(x, y) = f_X(x) f_Y(y) = e^{-x} \cdot 2e^{-2y} = 2e^{-(x+2y)}$ for $x \ge 0, y \ge 0$.
Calculating Conditional Probability
We need to find the conditional probability $P(X > 2Y \mid X > Y)$. Using the formula for conditional probability:
$ P(X > 2Y \mid X > Y) = \frac{P((X > 2Y) \cap (X > Y))}{P(X > Y)} $Note that if $X > 2Y$ and $Y \ge 0$, then $X$ is automatically greater than $Y$. Therefore, the intersection $(X > 2Y) \cap (X > Y)$ simplifies to just $X > 2Y$.
The formula becomes:
$ P(X > 2Y \mid X > Y) = \frac{P(X > 2Y)}{P(X > Y)} $Evaluating $P(X > Y)$
We calculate the probability $P(X > Y)$ by integrating the joint PDF over the region where $x > y$.
$ P(X > Y) = \int_{0}^{\infty} \int_{y}^{\infty} 2e^{-(x+2y)} dx dy $First, integrate with respect to $x$:
$ \int_{y}^{\infty} 2e^{-x} e^{-2y} dx = 2e^{-2y} \left[ -e^{-x} \right]_{y}^{\infty} = 2e^{-2y} (0 - (-e^{-y})) = 2e^{-3y} $Now, integrate the result with respect to $y$:
$ \int_{0}^{\infty} 2e^{-3y} dy = 2 \left[ -\frac{1}{3}e^{-3y} \right]_{0}^{\infty} = 2 \left( 0 - (-\frac{1}{3}) \right) = \frac{2}{3} $So, $P(X > Y) = \frac{2}{3}$.
Evaluating $P(X > 2Y)$
Next, we calculate the probability $P(X > 2Y)$ by integrating the joint PDF over the region where $x > 2y$.
$ P(X > 2Y) = \int_{0}^{\infty} \int_{2y}^{\infty} 2e^{-(x+2y)} dx dy $First, integrate with respect to $x$:
$ \int_{2y}^{\infty} 2e^{-x} e^{-2y} dx = 2e^{-2y} \left[ -e^{-x} \right]_{2y}^{\infty} = 2e^{-2y} (0 - (-e^{-2y})) = 2e^{-4y} $Now, integrate the result with respect to $y$:
$ \int_{0}^{\infty} 2e^{-4y} dy = 2 \left[ -\frac{1}{4}e^{-4y} \right]_{0}^{\infty} = 2 \left( 0 - (-\frac{1}{4}) \right) = \frac{2}{4} = \frac{1}{2} $So, $P(X > 2Y) = \frac{1}{2}$.
Final Calculation
Finally, we compute the conditional probability:
$ P(X > 2Y \mid X > Y) = \frac{P(X > 2Y)}{P(X > Y)} = \frac{1/2}{2/3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} $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 $1$?
- Consider a finite population of size $N = 100$. Let $T_1$ be the sample mean of a study variable based on a sample of size $n$ ($1 < n < N$) under simple random sampling with replacement scheme. Let $T_2$ be the sample mean of the same study variable based on a sample of size $n$ under simple random sampling without replacement scheme. If $Var(T_1) = 9Var(T_2)$, then the sample size $n$ equals
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 $\theta$?
