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# Lecture 24: Chapter 6 (Sections 1-3) ## The Poisson Distribution * **Definition**: A discrete random variable $X$ has a Poisson distribution with parameter $\lambda > 0$ if its probability mass function (PMF) is given by $$ P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!}, \quad k = 0, 1, 2...

# Lecture 24: Chapter 6 (Sections 1-3) ## The Poisson Distribution * **Definition**: A discrete random variable $X$ has a Poisson distribution with parameter $\lambda > 0$ if its probability mass function (PMF) is given by $$ P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!}, \quad k = 0, 1, 2, \dots $$ * **Properties**: * The mean of $X$ is $E[X] = \lambda$. * The variance of $X$ is $Var(X) = \lambda$. * The standard deviation of $X$ is $\sigma_X = \sqrt{\lambda}$. * **Example**: Suppose the number of cars arriving at a toll booth in one hour follows a Poisson distribution with a mean of 20 cars. What is the probability that exactly 15 cars arrive in one hour? * Here, $X \sim Poisson(20)$, so $\lambda = 20$. * We want to find $P(X=15)$. * Using the PMF: $$ P(X=15) = e^{-20} \frac{20^{15}}{15!} \approx 0.0516 $$ ## Continuous Random Variables * **Definition**: A continuous random variable $X$ is a random variable whose possible values can take on any value within a given range. * **Probability Density Function (PDF)**: The probability that $X$ falls within a certain interval $(a, b)$ is given by the integral of the PDF $f(x)$ over that interval: $$ P(a < X < b) = \int_a^b f(x) dx $$ * **Properties of PDF**: * $f(x) \geq 0$ for all $x$. * The total area under the PDF curve is equal to 1: $$ \int_{-\infty}^{\infty} f(x) dx = 1 $$ * **Cumulative Distribution Function (CDF)**: $$ F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) dt $$ * **Example**: Suppose $X$ has a PDF given by $$ f(x) = \begin{cases} 2x, & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases} $$ * Verify that $f(x)$ is a valid PDF: $$ \int_{-\infty}^{\infty} f(x) dx = \int_{0}^{1} 2x dx = [x^2]_0^1 = 1^2 - 0^2 = 1 $$ * Find $P(0.2 < X < 0.5)$: $$ P(0.2 < X < 0.5) = \int_{0.2}^{0.5} 2x dx = [x^2]_{0.2}^{0.5} = (0.5)^2 - (0.2)^2 = 0.25 - 0.04 = 0.21 $$ * Find $F(x)$: $$ F(x) = \int_{-\infty}^{x} f(t) dt = \int_{0}^{x} 2t dt = [t^2]_0^x = x^2, \quad 0 < x < 1 $$ Thus, $$ F(x) = \begin{cases} 0, & x \leq 0 \\ x^2, & 0 < x < 1 \\ 1, & x \geq 1 \end{cases} $$