Question

Let \(X\) and \(Y\) each take on either the value 1 or \(-1\). Let $$ \begin{aligned} p(1,1) &=P\\{X=1, Y=1\\} \\ p(1,-1) &=P[X=1, Y=-1\\} \\ p(-1,1) &=P[X=-1, Y=1\\} \\ p(-1,-1) &=P\\{X=-1, Y=-1\\} \end{aligned} $$ Suppose that \(E[X]=E[Y]=0\). Show that (a) \(p(1,1)=p(-1,-1) ;\) (b) \(p(1,-1)=p(-1,1)\). Let \(p=2 p(1,1) .\) Find (c) \(\operatorname{Var}(X)\); (d) \(\operatorname{Var}(Y)\) (e) \(\operatorname{Cov}(X, Y)\).

Short answer

The solution for the given problem is as follows: (a) \(p(1,1) = p(-1,-1)\) (b) \(p(1,-1) = p(-1,1)\) (c) \(\operatorname{Var}(X) = p\) (d) \(\operatorname{Var}(Y) = p\) (e) \(\operatorname{Cov}(X, Y) = 2p - 1\)

Step-by-step solution

Using the given expected values of X and Y.

Since we have \(E[X]=E[Y]=0\), we can use the definition of expectation to establish the following equations, \(E[X] = \sum_{x} x \cdot P[X=x] = 1 \cdot p(1,1) + 1 \cdot p(1,-1) - 1 \cdot p(-1,1) - 1 \cdot p(-1,-1) = 0\) \(E[Y] = \sum_{y} y \cdot P[Y=y] = 1 \cdot p(1,1) - 1 \cdot p(1,-1) + 1 \cdot p(-1,1) - 1 \cdot p(-1,-1) = 0\)

Proving the relations among joint probabilities.

Using the equations from Step 1, we can derive the relations required in parts (a) and (b): (a) Subtract the second equation from the first equation, we get \(2 \cdot p(1,1) - 2 \cdot p(-1,-1) = 0 \Rightarrow p(1,1) = p(-1,-1)\) (b) Add the second equation to the first equation, we get \(2 \cdot p(1,-1) - 2 \cdot p(-1,1) = 0 \Rightarrow p(1,-1) = p(-1,1)\)

Finding the probabilities using given value of p.

Using the given value of p, which is \(p = 2p(1,1)\), and using the relations derived in Step 2, we can find the joint probabilities: \(p(1,1) = \frac{p}{2}\) \(p(-1,-1) = \frac{p}{2}\) \(p(1,-1) = \frac{1-p}{2}\) \(p(-1,1) = \frac{1-p}{2}\)

Finding the variance for X and Y.

Now we can find the variance for both X and Y using the definition of variance and the joint probabilities we derived in Step 3: (c) \(\operatorname{Var}(X) = E[X^2] - (E[X])^2 = \sum_{x} x^2 \cdot P[X=x] = 1^2 \cdot p(1,1) + 1^2 \cdot p(1,-1) + (-1)^2 \cdot p(-1,1) + (-1)^2 \cdot p(-1,-1) = p\) (d) \(\operatorname{Var}(Y) = E[Y^2] - (E[Y])^2 = \sum_{y} y^2 \cdot P[Y=y] = 1^2 \cdot p(1,1) - 1^2 \cdot p(1,-1) + (-1)^2 \cdot p(-1,1) - (-1)^2 \cdot p(-1,-1) = p\)

Finding the covariance for X and Y.

Finally, we can find the covariance between X and Y using the definition of covariance and the joint probabilities we derived in Step 3: (e) \(\operatorname{Cov}(X, Y) = E[XY] - E[X] \cdot E[Y] = \sum_{x, y} (x \cdot y) \cdot P[X=x, Y=y] = 1 \cdot 1 \cdot p(1,1) + 1 \cdot (-1) \cdot p(1,-1) + (-1) \cdot 1 \cdot p(-1,1) + (-1) \cdot (-1) \cdot p(-1,-1) = p - (1-p) = 2p -1\)

Key concepts

Joint Probability Distribution

Understanding the joint probability distribution is crucial when studying two or more random variables simultaneously. It is the probability that each variable falls within a particular range of values. In this case, we are considering two discrete random variables, X and Y, which can each take on the value 1 or -1. The joint probabilities, such as p(1,1) or p(-1,-1), represent the chance that X and Y will take on their respective values at the same time.

For readability and to make sense of the data, it is often useful to arrange these probabilities into a matrix or table. This allows a more straightforward interpretation and calculation of any related statistics. When given certain conditions, like the expected values of X and Y being zero, it constrains our joint distribution in predictable ways, leading to relationships like p(1,1) = p(-1,-1) and p(1,-1) = p(-1,1). Such constraints help simplify the process of finding other probabilistic measures.

Probability Variance

Probability variance, represented as Var(X) for a random variable X, measures the spread of the random variable's probability distribution. It quantifies how much the values of the random variable deviate from the expected value or mean. A higher variance means the values are more spread out.

In the step-by-step solution, we calculated Var(X) and Var(Y) after determining the joint probabilities. This was done using the formula Var(X) = E[X^2] - (E[X])^2, which reflects the average squared deviation of the random variable's outcomes from its mean. Since X and Y have equal variance in this example, it simplifies comparisons and further calculations between these two variables.

Probability Covariance

Probability covariance, noted as Cov(X, Y), is a measure that indicates the extent to which two random variables change together. If X and Y tend to show similar behavior (both increase or decrease together), the covariance is positive; if they tend to show opposite behavior (one increases while the other decreases), the covariance is negative. Zero covariance suggests no linear relationship between the variables.

The formula used to find covariance is Cov(X, Y) = E[XY] - E[X]E[Y]. The exercise shows how to apply this formula by summing over all combinations of X and Y, weighted by their joint probabilities. Covariance is vital in finance and other fields for understanding how different variables influence each other.

Expected Value of Random Variables

The expected value of a random variable gives us a 'central' or 'average' value that the variable is expected to take. Mathematically expressed as E[X], it's computed by summing all possible values, each multiplied by its probability of occurring.

This concept is foundational to statistics and probability, as it provides the basis for more complex measures like variance and covariance. In our example, since the expected values E[X] and E[Y] are both zero, it significantly simplified the calculations and showed that neither random variable is expected to deviate from zero on average. This property greatly assists in understanding the behavior of probabilistic models.