Question
Let $X$ and $Y$ be independent exponential random variables. If $E[X] = 1$ and $E[Y] = \frac{1}{2}$ then $P(X > 2Y \mid X > Y)$ is
The correct answer is
$\frac{3}{4}$
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} $Was this answer helpful?
Important Questions from Random Variables
- A mobile manufacturing company uses two brands of batteries for its mobiles. The life (in years) of batteries of Brand I follows an exponential distribution with the probability density function
$ f(x) = \begin{cases} e^{-x}, & \text{if } x>0, \\ 0, & \text{otherwise,} \end{cases} $
and that of Brand II follows a gamma distribution with the probability density function
$ g(x) = \begin{cases} \frac{x}{4} e^{-x/2}, & \text{if } x>0, \\ 0, & \text{otherwise.} \end{cases} $
The company uses the batteries of Brands I and II in proportion of $20\%$ and $80\%$ respectively, in its mobiles. The probability that a randomly selected mobile has the battery life more that $2$ years is - Consider a discrete random variable $X$ with the probability mass function
$ P(X = 0) = \frac{\theta}{3}, \ P(X = 1) = 1 - \frac{\theta}{2}, \ P(X = 2) = \frac{\theta}{6}, $
where $\theta \in (0,1)$ is an unknown parameter. In a random sample of size $90$ from this distribution, the observed counts for $X = 0, 1$ and $2$ are $20, 60$ and $10$, respectively. Then, the maximum likelihood estimate of $\theta$ is - Let $X$ be a random sample of size $1$ from the probability density function
$ f(x|\theta) = \begin{cases} \frac{3}{\theta^3} (\theta - x)^2, & \text{if } 0<x<\theta, \\ 0, & \text{otherwise.} \end{cases} $
If $ \left(\frac{X}{1-\lambda_1}, \frac{X}{1-\lambda_2}\right) $ is a confidence interval for $\theta$ with confidence coefficient $1 - \alpha$, where $\lambda_i \in (0,1), \ i = 1,2, \ \lambda_1<\lambda_2$, and $\alpha \in (0,1)$, then which of the following statements is true? - Let $X_1, X_2, . . ., X_n$ be a random sample from a continuous distribution with the common probability density function
$ f(x|\theta) = \begin{cases} \frac{2\theta^2}{x^{\theta+1}}, & \text{if } x>2, \\ 0, & \text{otherwise,} \end{cases} $
where $\theta (> 0)$ is an unknown parameter. Suppose $P(Y>\chi^2_{m,\beta}) = \beta$, where $Y \sim \chi^2_m$. For testing $H_0: \theta = 1$ against $H_1 : \theta>1$, a uniformly most powerful test of size $\alpha, \ 0<\alpha<1$, will reject $H_0$ if - Suppose we want to estimate the population mean $\bar{Y}$ of a variable for a finite population of size $85$, with $34$ Statisticians and $51$ Biologists. We consider the following sampling scheme:
A stratified random sample with $2$ strata of Statisticians (Stratum-1) and Biologists (Stratum-2), where $12$ Statisticians and $15$ Biologists are drawn from Stratum-1 and Stratum-2, respectively, using SRSWOR scheme.
Denote $\bar{y}_S, \bar{y}_B$, and $\bar{y}$ as the mean of the variable among the Statistician sample, Biologist sample, and the combined sample, respectively. Which of the following is an unbiased estimator of $\bar{Y}$?
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