Problem 74 - Calculus Single Variable

Question

The standard normal probability density \(f\) satisfies the differential equation $$ f^{\prime}(x)=-x f(x), f(0)=\frac{1}{\sqrt{2 \pi}} $$ Find \(f(x)\)

Step-by-Step Solution

Answer

The function \( f(x) \) is \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \).

Step 1: Understanding the Differential Equation

The question provides a differential equation for the standard normal probability density function, which is given by \( f'(x) = -x f(x) \). We also have an initial condition \( f(0) = \frac{1}{\sqrt{2\pi}} \). The goal is to find \( f(x) \).

Step 2: Solving the Differential Equation

We first rearrange the differential equation into a separable form: \( \frac{f'(x)}{f(x)} = -x \). We can integrate both sides with respect to \( x \) to find \( f(x) \).

Step 3: Integration

Integrating the left side \( \int \frac{f'(x)}{f(x)} \, dx = \int -x \, dx \) gives us \( \ln |f(x)| = -\frac{x^2}{2} + C \) where \( C \) is the constant of integration.

Step 4: Solving for f(x)

Exponentiating both sides to solve for \( f(x) \) gives \( |f(x)| = e^{-x^2/2 + C} = Ce^{-x^2/2} \). Since \( f(x) \) represents a probability density, it must be positive, so \( f(x) = Ce^{-x^2/2} \).

Step 5: Applying the Initial Condition

Using the given initial condition \( f(0) = \frac{1}{\sqrt{2\pi}} \), we substitute \( x = 0 \) into our expression for \( f(x) \) to find \( C \): \( f(0) = Ce^{0} = C = \frac{1}{\sqrt{2\pi}} \).

Step 6: Conclusion

By substituting back \( C = \frac{1}{\sqrt{2\pi}} \) into the expression for \( f(x) \), we find \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \). This is the standard normal probability density function.

Key Concepts

Differential Equation

A differential equation involves functions and their derivatives. It's like a mathematical recipe describing how a function changes. In this exercise, we have the differential equation \( f'(x) = -x f(x) \). It tells us how the slope of the function, \( f(x) \), behaves as \( x \) changes. Here, \( f'(x) \) represents the derivative or 'rate of change' of the probability density function.

• Derivatives show how steep a curve is at any point.
• Differential equations can be used to model real-world phenomena, like population growth or motion.

For our exercise, solving the differential equation means finding a function \( f(x) \) that satisfies this rate of change description. Using the initial condition \( f(0) = \frac{1}{\sqrt{2\pi}} \) helps fine-tune our solution, giving a complete answer.

Standard Normal Distribution

The standard normal distribution is a special kind of probability distribution. It's centered at zero and has a standard deviation of one. We often use it in statistics because it's well-behaved and describes many natural phenomena.

• The curve is symmetric about zero.
• Most values (about 68%) lie within one standard deviation from the mean.
• The equation for the probability density function is \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \).

This function is what we solved for in the exercise. It's called the probability density function (PDF) because it shows how likely different outcomes are. In a broader sense, the standard normal distribution is a building block for other statistical analyses.

Probability Theory

Probability theory is the study of randomness and uncertainty. It's a mathematical framework for predicting the likelihood of events. Probability density functions, like the one in this exercise, play a crucial role.

• Probability values range from 0 (impossible event) to 1 (certain event).
• For continuous outcomes, the PDF helps us understand where values are likely to fall.
• A key feature of a valid PDF: the total area under the curve equals 1, representing total certainty that some outcome will happen.

In probability theory, the standard normal distribution is widely used to model uncertainty. It simplifies many problems by allowing us to use standardized scores, which make data easier to handle.

Separation of Variables

Separation of variables is a technique to solve differential equations. It involves rearranging the equation so that each side depends on only one variable. This makes it easier to integrate both sides independently.In solving \( f'(x) = -x f(x) \), we transformed it to \( \frac{f'(x)}{f(x)} = -x \). This separation allowed us to integrate each side with respect to \( x \), leading to a solution for \( f(x) \).

• Separating variables simplifies the solving process.
• It requires an equation where variables can be cleanly split between sides.
• It's often a first step in more complex differential equation techniques.

This method is especially useful for problems like this one, where we can express the relationship in terms of a constant of integration—a crucial step in finding the probability density function.