Question

Let \(X\) and \(Y\) have the joint pmf \(p(x, y)=e^{-2} /[x !(y-x) !], y=0,1,2, \ldots\), \(x=0,1, \ldots, y\), zero elsewhere. (a) Find the mgf \(M\left(t_{1}, t_{2}\right)\) of this joint distribution. (b) Compute the means, the variances, and the correlation coefficient of \(X\) and \(Y\). (c) Determine the conditional mean \(E(X \mid y)\). Hint: Note that $$ \sum_{x=0}^{y}\left[\exp \left(t_{1} x\right)\right] y ! /[x !(y-x) !]=\left[1+\exp \left(t_{1}\right)\right]^{y} $$ Why?

Short answer

The calculations comprise multiple steps. Step 1: Calculate mgf \(M(t1, t2)\). Step 2: Calculate the mean, variance and correlation for X and Y. Step 3: Calculate the conditional mean \(E(X|Y=y)\). This process involves a significant amount of calculations.

Step-by-step solution

Calculation of mgf \(M(t_{1}, t_{2})\)

To find the moment generating function (mgf) we use the formula \(M(t1, t2) = E[e^{t1X +t2Y}]\). Here, we substitute the given pmf and sum it over all the values of \(x\) and \(y\) within the defined range, resulting in \(M(t1, t2) = \sum_{y=0}^{\infty} \sum_{x=0}^{y} e^{t1x + t2y} p(x,y)\) which simplifies to \(M(t1, t2) = e^{-2} \sum_{y=0}^{\infty} e^{t2y}y! \sum_{x=0}^{y} \frac{e^{t1x}}{x!(y-x)!}\)

Compute means, variances and correlation

Next step is to compute mean, variance and correlation. The mean of \(X\) and \(Y\) can be calculated using their respective mgf's as \(\mu_X = M'(0)\) and \(\mu_Y = M'(0)\). Similarly, variances can be calculated as \(\sigma_X^2 = M''(0) - [M'(0)]^2\) and \(\sigma_Y^2 = M''(0) - [M'(0)]^2\). After calculating these, correlation coefficient can be found using the formula: \[ \rho_{X, Y} = \frac{Cov(X, Y)}{\sigma_X \sigma_Y} \] where \(Cov(X, Y) = E[XY] - \mu_X \mu_Y\).

Calculate Conditional mean \(E(X|y)\)

The conditional expectation of \(X\) given \(Y=y\) can be calculated using the formula: \[E(X|Y=y) = \sum_{x=0}^{y} x p(x|Y=y)\] \[p(x|Y=y)\] can be found by dividing the joint pmf by the marginal pmf of \(Y\) which can be calculated as \[p_Y(y) = \sum_{x=0}^{y} p(x, y)\]. First calculate \(p_Y(y)\) and then \(p(x|Y=y)\) and substitute these in the formula for \(E(X|Y=y)\).

Key concepts

Moment Generating Function

The moment generating function (mgf) is a powerful tool that encapsulates all the moments (e.g. mean, variance) of a probability distribution. Think of it as a machine that produces the 'moments' of a distribution on demand. For a joint probability distribution of random variables, like in our textbook exercise, the mgf is given by the expectation value of the exponential of a linear combination of those variables:
In simpler terms, for variables X and Y with a joint probability mass function (pmf) p(x, y), their joint mgf is defined by \[ M(t_1, t_2) = E[e^{t_1X + t_2Y}] \].
For our exercise, this requires summing over all valid values of X and Y while using the given pmf. Calculating this not only helps in deriving moments, but also offers insights into the dependence structure between X and Y.
Through this step by step solution, one can find that the mgf has the unexpected ability to simplify complex summations, as seen in the hint provided.

Mean and Variance of a Distribution

The mean and the variance are fundamental characteristics of any probability distribution.

• The mean, often denoted by \( \mu \), is the average or expected value of the distribution.
• The variance, shown as \( \sigma^2 \), measures the spread or dispersion of the values in that distribution.

For continuous distributions, these can be found by integrating; for discrete distributions, this transitions to summation.
Our exercise involves calculating these parameters for both X and Y. Using the mgf, we can extract the mean by taking the first derivative of the mgf with respect to \( t \) and evaluating it at 0 (i.e., \( M'(0) \)).
The variance involves a bit more calculus - it's found by taking the second derivative of the mgf (i.e., \( M''(0) \)) and then subtracting the square of the mean. This is very much like peeling layers off an onion; each step lets us delve deeper into the data's core properties. By following this methodology, we can understand the essential nature of random variables in our exercise.

Correlation Coefficient

The correlation coefficient, often symbolized as \( \rho \), measures the strength and direction of a linear relationship between two variables. It can range from -1 to 1, where -1 implies perfect negative linear correlation, 1 indicates perfect positive linear correlation, and 0 signifies no linear correlation.
In our textbook problem, we're not just looking at random variables in isolation; we want to understand how X and Y move together. To ascertain this, we compute the correlation coefficient by dividing the covariance of X and Y by the product of their standard deviations. It involves the expected product of these variables, subtracting the product of their means. This computation helps to draw conclusions about whether knowing one variable provides any information about the other.
The solution approach provides us with a stepwise path to the correlation coefficient. Once the means and variances are established from using the mgf, the expectation of the product of X and Y leads us to the covariance needed for our final calculation. It's like piecing together a puzzle - combining the result from the means and variances to get a complete picture of the relationship between X and Y.