Understanding the Joint Probability Density

Oct 13, 2025 by magicalceylontours.com 125 views

A1. Computing the Variance of a Poisson Variable with Uniform Mean

In this problem, we are given a Poisson variable $X$ with mean $\lambda$ , where $\lambda$ is uniformly distributed on the interval $[-2,2]$ . Our goal is to compute the variance of $X$ , denoted as $\operatorname{Var}(X)$ .

Understanding the Distribution of $\lambda$

The mean of a Poisson variable is given by $\lambda$ , and in this case, $\lambda$ is uniformly distributed on the interval $[-2,2]$ . This means that the probability density function (pdf) of $\lambda$ is given by:

f(\lambda) = \frac{1}{4} \quad \text{for} \quad -2 \leq \lambda \leq 2

Computing the Variance of $X$

To compute the variance of $X$ , we need to use the formula:

\operatorname{Var}(X) = E(X^2) - (E(X))^2

where $E(X)$ is the expected value of $X$ and $E(X^2)$ is the expected value of $X^2$ .

Expected Value of $X$

Since $X$ is a Poisson variable with mean $\lambda$ , we know that:

E(X) = \lambda

Expected Value of $X^2$

To compute the expected value of $X^2$ , we need to use the formula:

E(X^2) = E(\lambda^2) + 2E(\lambda)

where $E(\lambda^2)$ is the expected value of $\lambda^2$ and $E(\lambda)$ is the expected value of $\lambda$ .

Computing $E(\lambda^2)$

Since $\lambda$ is uniformly distributed on the interval $[-2,2]$ , we can compute the expected value of $\lambda^2$ as follows:

E(\lambda^2) = \int_{-2}^{2} \lambda^2 \cdot \frac{1}{4} d\lambda = \frac{1}{4} \int_{-2}^{2} \lambda^2 d\lambda

Evaluating the integral, we get:

E(\lambda^2) = \frac{1}{4} \left[ \frac{\lambda^3}{3} \right]_{-2}^{2} = \frac{1}{4} \left( \frac{8}{3} - \frac{-8}{3} \right) = \frac{16}{9}

Computing $E(\lambda)$

Since $\lambda$ is uniformly distributed on the interval $[-2,2]$ , we can compute the expected value of $\lambda$ as follows:

E(\lambda) = \int_{-2}^{2} \lambda \cdot \frac{1}{4} d\lambda = \frac{1}{4} \int_{-2}^{2} \lambda d\lambda

Evaluating the integral, we get:

E(\lambda) = \frac{1}{4} \left[ \frac{\lambda^2}{2} \right]_{-2}^{2} = \frac{1}{4} \left( 2 - 2 \right) = 0

Computing the Variance of $X$

Now that we have computed the expected value of $X$ and the expected value of $X^2$ , we can compute the variance of $X$ as follows:

\operatorname{Var}(X) = E(X^2) - (E(X))^2 = E(\lambda^2) + 2E(\lambda) - (E(\lambda))^2

Substituting the values we computed earlier, we get:

\operatorname{Var}(X) = \frac{16}{9} + 2(0) - (0)^2 = \frac{16}{9}

Therefore, the variance of $X$ is $\frac{16}{9}$ .

In this problem, we computed the variance of a Poisson variable $X$ with mean $\lambda$ , where $\lambda$ is uniformly distributed on the interval $[-2,2]$ . We used the formula for the variance of a Poisson variable and computed the expected value of $X$ and the expected value of $X^2$ . We then used these values to compute the variance of $X$ , which is $\frac{16}{9}$ .

In this problem, we are given two random variables $X$ and $Y$ with joint probability density function (pdf) $f(x,y)$ . Our goal is to compute the marginal pdfs of $X$ and $Y$ .

Understanding the Joint Probability Density

The joint pdf of $X$ and $Y$ is given by:

f(x,y) = \begin{cases} 2xy & \text{if} \quad 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}

Computing the Marginal Pdf of $X$

To compute the marginal pdf of $X$ , we need to integrate the joint pdf over all possible values of $Y$ . This gives us:

f_X(x) = \int_{-\infty}^{\infty} f(x,y) dy = \int_{0}^{1} 2xy dy

Evaluating the integral, we get:

f_X(x) = \left[ xy^2 \right]_{0}^{1} = x

Computing the Marginal Pdf of $Y$

To compute the marginal pdf of $Y$ , we need to integrate the joint pdf over all possible values of $X$ . This gives us:

f_Y(y) = \int_{-\infty}^{\infty} f(x,y) dx = \int_{0}^{1} 2xy dx

Evaluating the integral, we get:

f_Y(y) = \left[ x^2y \right]_{0}^{1} = y

Q1: What is the Poisson distribution?

A1: The Poisson distribution is a discrete probability distribution that models the number of events occurring in a fixed interval of time or space, where these events occur with a known constant mean rate and independently of the time since the last event.

Q2: What is the formula for the Poisson distribution?

A2: The formula for the Poisson distribution is:

P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}

where $k$ is the number of events, $\lambda$ is the mean rate of events, and $e$ is the base of the natural logarithm.

Q3: What is the variance of a Poisson distribution?

A3: The variance of a Poisson distribution is equal to its mean, $\lambda$ . This means that the variance of a Poisson distribution is $\lambda$ .

Q4: What is the joint probability density function?

A4: The joint probability density function (pdf) of two random variables $X$ and $Y$ is a function that describes the probability of different combinations of values of $X$ and $Y$ .

Q5: How do you compute the marginal pdf of a random variable?

A5: To compute the marginal pdf of a random variable, you need to integrate the joint pdf over all possible values of the other variable.

Q6: What is the formula for the marginal pdf of a random variable?

A6: The formula for the marginal pdf of a random variable is:

f_X(x) = \int_{-\infty}^{\infty} f(x,y) dy

where $f(x,y)$ is the joint pdf of $X$ and $Y$ .

Q7: What is the difference between the joint pdf and the marginal pdf?

A7: The joint pdf describes the probability of different combinations of values of two random variables, while the marginal pdf describes the probability of different values of one random variable.

Q8: How do you use the joint pdf to compute the marginal pdf?

A8: To use the joint pdf to compute the marginal pdf, you need to integrate the joint pdf over all possible values of the other variable.

Q9: What is the relationship between the joint pdf and the marginal pdf?

A9: The joint pdf can be used to compute the marginal pdf by integrating over all possible values of the other variable.

Q10: What is the importance of the joint pdf and the marginal pdf in probability theory?

A10: The joint pdf and the marginal pdf are important concepts in probability theory because they allow us to describe the probability of different combinations of values of random variables and to compute the probability of different values of one random variable.

In this article, we have answered some common questions about the Poisson distribution and joint probability densities. We have discussed the formula for the Poisson distribution, the variance of a Poisson distribution, the joint probability density function, and the marginal pdf of a random variable. We have also discussed the relationship between the joint pdf and the marginal pdf and the importance of these concepts in probability theory.