Question

Let \(X\) and \(Y\) be jointly distributed discrete random variables having possible values \(0,1,2, \ldots\) For \(|s|<1,|t|<1\) define the joint generating function $$ \phi_{X, y}(s, t)=\sum_{i, j=0}^{\infty} s^{i} t^{j} \operatorname{Pr}\\{X=i, Y=j\\} $$ and the marginal generating functions $$ \begin{aligned} &\phi_{X}(s)=\sum_{i=0}^{\infty} s^{i} \operatorname{Pr}\\{X=i\\} \\ &\phi_{Y}(t)=\sum_{j=0}^{\infty} t^{j} \operatorname{Pr}\\{Y=j\\} \end{aligned} $$ (a) Prove that \(X\) and \(Y\) are independent if and only if $$ \phi_{X, y}(s, t)=\phi_{X}(s) \phi_{Y}(t) \quad \text { for all } s, t $$ (b) Give an example of jointly distributed random variables \(X, Y\) which. are not independent, but for which. $$ \phi_{X, Y}(t, t)=\phi_{X}(t) \phi_{Y}(t) \text { for all } t $$ (This example is pertinent because \(\phi_{X, Y}(t, t)\) is the generating function of the sum \(X+Y\). Thus independence is sufficient but not necessary for the generating function of a sum of random variables to be the product of the marginal generating functions.)

Short answer

In summary, we have shown that if the joint generating function of two discrete random variables X and Y equals the product of their marginal generating functions, i.e., \(\phi_{X, Y}(s, t) = \phi_{X}(s) \phi_{Y}(t)\) for all \(s, t\), then X and Y are independent. Conversely, if X and Y are independent, their joint generating function is equal to the product of their marginal generating functions. We also provided an example of non-independent random variables satisfying the condition \(\phi_{X, Y}(t, t) = \phi_{X}(t) \phi_{Y}(t)\) for all \(t\).

Step-by-step solution

If X and Y are independent, then the probability of their joint events can be written as the product of their individual probabilities, i.e., \(\operatorname{Pr}\\{X=i, Y=j\\} = \operatorname{Pr}\\{X=i\\} \cdot \operatorname{Pr}\\{Y=j\\}\). Now, let's plug this into the joint generating function: \[ \begin{aligned} \phi_{X, Y}(s, t) &= \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} s^{i} t^{j} \operatorname{Pr}\\{X=i, Y=j\\} \\ &= \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} s^{i} t^{j} (\operatorname{Pr}\\{X=i\\} \cdot \operatorname{Pr}\\{Y=j\\}) \\ \end{aligned} \] #Step 2: Separate the sum and rewrite it as the product of marginal generating functions#

We can separate the sum as follows: \[ \begin{aligned} \phi_{X, Y}(s, t) &= \sum_{i=0}^{\infty} s^{i} \operatorname{Pr}\\{X=i\\} \sum_{j=0}^{\infty} t^{j} \operatorname{Pr}\\{Y=j\\} \\ &= \phi_{X}(s) \phi_{Y}(t) \end{aligned} \] So, if X and Y are independent, \(\phi_{X, Y}(s, t) = \phi_{X}(s) \phi_{Y}(t) \) for all \(s, t\). #Step 3: Prove the converse, i.e., if the joint generating function equals the product of the marginal generating functions, X and Y are independent#

Assuming \(\phi_{X, Y}(s, t) = \phi_{X}(s) \phi_{Y}(t)\) for all \(s, t\), we have: \[ \begin{aligned} \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} s^{i} t^{j} \operatorname{Pr}\\{X=i, Y=j\\} &= \left(\sum_{i=0}^{\infty} s^{i} \operatorname{Pr}\\{X=i\\}\right) \left(\sum_{j=0}^{\infty} t^{j} \operatorname{Pr}\\{Y=j\\}\right) \end{aligned} \] Since this equality holds for all \(s, t\), we can equate the coefficients of the powers of \(s^i t^j\): \[ \operatorname{Pr}\\{X=i, Y=j\\} = \operatorname{Pr}\\{X=i\\} \cdot \operatorname{Pr}\\{Y=j\\} \] Thus, X and Y are independent. #Step 4: Provide an example of non-independent random variables satisfying the given condition#

Let's consider the following jointly distributed random variables X and Y: \( \begin{array}{c|c|c} \text{X} & \text{Y}\\ \hline 1 & 1 \\ 1 & -1 \\ -1 & 1 \\ -1 & -1 \\ \end{array} \) Each of these combinations has a probability of \(\dfrac{1}{4}\). Let's compute the joint generating function: \[ \phi_{X, Y}(t, t) = \dfrac{1}{4}(t^2 + t^2 + t^2 + t^2) = t^2 \] Now, let's compute the marginal generating functions: \[ \phi_{X}(t) = \dfrac{1}{2}(t + t^{-1}) \quad \text{and} \quad \phi_{Y}(t) = \dfrac{1}{2}(t + t^{-1}) \] So, for all \(t\), \[ \phi_{X, Y}(t, t) = \phi_{X}(t) \phi_{Y}(t) \] However, X and Y are not independent because their joint distribution is not a product of their marginals, i.e., \(\operatorname{Pr}\\{X=1, Y=1\\} = \dfrac{1}{4} \neq \dfrac{1}{2} \cdot \dfrac{1}{2}\).