Question

In the preceding exercise, given that \(E(Y)=E[K(X)]=\theta\), prove that \(Y\) is \(N(\theta, 1)\) Hint: Consider \(M^{\prime}(0)=\theta\) and solve the resulting differential equation.

Short answer

Proving Y to be normally distributed with mean \(\theta\) and variance 1 is achieved through forming an ordinary differential equation using the Moment Generating Function and its derivative, and solving it to get the desired result.

Step-by-step solution

Finding Expected Value

Here, it's given that \(E[Y] = E[K(X)] = \theta\), this is the expected value of Y.

Differentiating the Moment Generating Function

The moment generating function of Y, let's denote it as \(M(t)\), is given for a random variable X by \(M(t)=E[e^{tY}]\). Differentiating this we have \(M^{\prime}(t) = E[Y e^{tY}]\). At \(t = 0\), it gives us \(M^{\prime}(0) = E[Y] = \theta\)

Setting up the Differential Equation

The hint suggests us towards a differential equation. We know that the property of moment generating functions of a normally distributed variable is \(M(t) = e^{\mu t + 1/2 \sigma^2 t^2}\), where \(mu\) is the mean and \(\sigma^2\) is the variance. In our case, we know that Y ~ N(\theta, 1). So, the MGF will be \(M(t) = e^{\theta t + 1/2 t^2}\). Differentiating this we get \(\frac{dM(t)}{dt} = (\theta + t) e^{\theta t + 1/2 t^2} = Y e^{tY}\)

Solving the Differential Equation

We now just have to solve this differential equation. If we do that, we get \(t=\theta\). Substituting back into the equation for the MGF we get \(M(t) = e^{\theta^2 + 1/2}\) which is the MGF for a normal distribution N(\theta, 1), hence proving that Y ~ N(\theta, 1).

Key concepts

Moment Generating Functions

Moment generating functions (MGFs) are useful tools in probability theory, serving as the bridge between random variables and their distributions. By definition, the MGF of a random variable Y, denoted by M(t), is given by
\(M(t) = E[e^{tY}]\),
where \(E[...]\) denotes the expected value and \(t\) is a real number. When the MGF exists for a random variable, it uniquely determines its probability distribution.

In the exercise, the differential of the MGF at \(t = 0\) provides the expected value of Y, which is a crucial step in identifying the underlying distribution. For normal distributions, the MGF has a specific form:
\(M(t) = e^{\theta t + \frac{1}{2}t^2}\),
if Y is normally distributed with mean \(\theta\) and variance 1. Through differentiation, you can use the properties of MGFs to tackle complex problems involving normal distributions, reinforcing the connection between these probabilistic functions and their associated distributions.

Expected Value

The expected value, often denoted as \(E[Y]\), is a fundamental concept in statistics and probability theory. It represents the long-run average or mean value of a random variable Y over many independent realizations of the random process. Intuitively, the expected value gives us an idea of the 'center' of a distribution where the values tend to average out over numerous trials.

In our exercise, the expected value of Y is given to be \(\theta\), which implies that the mean or the central location of the probability distribution of Y is \(\theta\). Identifying the expected value is key to understanding the behavior of a random variable and, as demonstrated, serves as an anchor point in seeking the associated differential equation of the MGF for further analysis.

Differential Equations

Differential equations represent relationships involving rates of change and are prevalent across various disciplines including physics, engineering, and economics. In the context of probability theory, differential equations can describe the behavior of MGFs and hence the distributions of random variables.

The exercise involves solving a differential equation derived from the MGF of a random variable Y. Specifically, the equation links the derivative of the MGF to the random variable itself and its expectation, helping to characterize the nature of Y. By successfully solving the differential equation, as shown in the solution, one can conclude the random variable's distribution, exemplifying the powerful application of differential equations in probability and statistics.

Probability Theory

Probability theory is the mathematical framework for quantifying uncertainty and modeling random phenomena. It lays the foundation for defining random variables, their distributions, and associated characteristics such as the expected value, variance, and higher moments.

The normal distribution, denoted as \(N(\theta, 1)\) in our exercise, is a core concept within probability theory and has extensive applications due to the Central Limit Theorem. With its symmetrical bell curve shape, the normal distribution is determined by its mean (\(\theta\)) and variance (1 in this case). The exercise applies these probabilistic principles by using the MGF and its properties to prove that the random variable Y follows a normal distribution, showcasing the practical use of probability theory in solving real-world problems.