1. **Problem Statement:** Find the PDF, expected value, and variance of an exponential random variable $X \sim \text{Exponential}(\lambda)$. 2. **Formula for Exponential Distribution:** - PDF: $f_X(x) = \lambda e^{-\lambda x}$ for $x \geq 0$ - Expected value: $E[X] = \frac{1}{\lambda}$ - Variance: $\text{Var}(X) = \frac{1}{\lambda^2}$ 3. **Explanation:** - The exponential distribution models the time between events in a Poisson process. - The PDF decreases exponentially from $x=0$. - The mean is the reciprocal of the rate $\lambda$. - The variance is the square of the mean. 4. **Intermediate Work:** - PDF is given by definition. - Expected value calculation: $$E[X] = \int_0^\infty x \lambda e^{-\lambda x} dx = \frac{1}{\lambda}$$ - Variance calculation: $$E[X^2] = \int_0^\infty x^2 \lambda e^{-\lambda x} dx = \frac{2}{\lambda^2}$$ $$\text{Var}(X) = E[X^2] - (E[X])^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}$$ --- 1. **Problem Statement:** Find the PDF, expected value, and variance of a normal random variable $Z \sim \mathcal{N}(\mu, \sigma^2)$. 2. **Formula for Normal Distribution:** - PDF: $$f_Z(z) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(z-\mu)^2}{2\sigma^2}}$$ - Expected value: $E[Z] = \mu$ - Variance: $\text{Var}(Z) = \sigma^2$ 3. **Explanation:** - The normal distribution is symmetric and bell-shaped. - The mean $\mu$ is the center of the distribution. - The variance $\sigma^2$ controls the spread. 4. **Intermediate Work:** - PDF is defined by the formula above. - Expected value and variance are parameters of the distribution. **Final answers:** - Exponential: $f_X(x) = \lambda e^{-\lambda x}, E[X] = \frac{1}{\lambda}, \text{Var}(X) = \frac{1}{\lambda^2}$ - Normal: $f_Z(z) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(z-\mu)^2}{2\sigma^2}}, E[Z] = \mu, \text{Var}(Z) = \sigma^2$