Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\mu, \sigma^{2}\right)\). (a) If the constant \(b\) is defined by the equation \(\operatorname{Pr}(X \leq b)=0.90\), find the mle of \(b\). (b) If \(c\) is given constant, find the mle of \(\operatorname{Pr}(X \leq c)\).

Short answer

The maximum likelihood estimate of \(b\) is the sample mean. The maximum likelihood estimate of \(Pr(X \leq c)\) is \(\Phi((c-\mu)/\sigma)\), where \(\mu\) and \(\sigma\) are the sample mean and sample standard deviation respectively.

Step-by-step solution

Calculate Pr(X ≤ b) when μ = b

The cumulative distribution function of a standard normal random variable \(Z\) is given by \(\Phi(Z)\). For a normally distributed variable \(X\) with mean \(\mu\) and variance \(\sigma^2\), the standardization is \(Z = (X-\mu)/\sigma\). Hence, \(Pr(X \leq b) = \Phi((b-\mu)/\sigma)\). But the problem states that \(Pr(X \leq b) = 0.90\), so \(\Phi((b-\mu)/\sigma) = 0.90\). If we let \(\mu = b\) (since we are trying to find the maximum likelihood estimate of \(b\)), then \(\Phi((b-b)/\sigma) = 0.90\), which simplifies to \(\Phi(0) = 0.90\).

Determine the value of b from the equation

\(\Phi(0) = 0.5\) for any normal distribution. Because \(0.9 > 0.5\), the value of \(b\) must be greater than \(\mu\). Since we have chosen \(\mu\) to be equal to \(b\), the maximum likelihood estimate of \(b\) is the same as the maximum likelihood estimate of \(\mu\) for the normal distribution \(N(\mu, \sigma^{2})\), which is the sample mean.

Calculate Pr(X ≤ c) and find the mle

For question part (b), we are trying to find the maximum likelihood estimate of the cumulative probability \(Pr(X \leq c)\) for a given constant \(c\). This probability can be calculated as \(\Phi((c-\mu)/\sigma)\), using the properties of the normal distribution. Again, replacing \(\mu\) with the sample mean and \(\sigma\) with the sample standard deviation gives the maximum likelihood estimate of the cumulative probability.

Key concepts

Normal Distribution

Imagine you're looking at a bell-shaped curve where most of the data cluster around a central point, with fewer observations as you move away from the center. This is a normal distribution, a fundamental concept in statistics, representing a continuous probability distribution with a symmetrical shape. The beauty of the normal distribution is its predictability - about 68% of data falls within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three.

In our exercise, we see a variable X that is normally distributed, written as N(μ, σ²), with μ representing the mean (the peak's location on the horizontal axis) and σ² denoting the variance (which determines the spread or width of the distribution).

Understanding the characteristics of a normal distribution is vital when performing maximum likelihood estimation (MLE), since it allows for parameter estimation, like finding the mean – the most likely place to find our observations.

Cumulative Distribution Function

Diving deeper, the cumulative distribution function (CDF), represented by Pr(X ≤ x), tells us the probability that a random variable X will take a value less than or equal to x. It's essentially adding up probabilities from '-infinity' to 'x', giving us insights into the likelihood of various outcomes.

In the problem, we use the CDF of the normal distribution to find out where to draw the line so that 90% of our data falls below it. This function steadily increases from 0 to 1 as you move from left to right across the distribution.

The exercise asks for the maximum likelihood estimate (MLE) of b when the CDF is set to 0.90. We can say that the value of b, in this case, divides the area under the curve such that 90% of the total area is to the left of b. The CDF plays a crucial role in MLE as it helps locate the threshold for the desired probability, allowing for informed decisions based on statistical evidence.

Standard Normal Variable

Now, let's introduce the concept of a standard normal variable, often symbolized as Z. A standard normal distribution is a special case of the normal distribution where the mean (μ) is 0 and the variance (σ²) is 1. This Z variable is fundamental because it helps us standardize any normal distribution so that we can use published tables or software to find probabilities and percentiles.

In our MLE problem, we standardize our normal variable X to become a Z by using the formula Z = (X - μ) / σ. This method allows us to compare different normal distributions and to use the known properties of the standard normal distribution to find probabilities like Pr(X ≤ b) = Φ((b - μ) / σ) as given in the step-by-step solution.

Understanding the transformation to a standard normal variable simplifies calculations and makes it easier to determine probabilities for any normally distributed variable. This concept is central to many statistical methods, including hypothesis testing and confidence interval estimation.