# Need help with expectation value

• begyu85
In summary, the conversation discusses calculating the expectation value of the products of a random, uniformly distributed vector's Cartesian components. The suggested method involves converting to spherical coordinates, but the correct solution can be found without this conversion. The integral should also be multiplied by the probability density function, f(x), which was not accounted for in the attempt at a solution. Additionally, the value of the integral is 1/3, but the incorrect result of 1/9 was obtained due to a missing factor of f(x)^2.
begyu85

## Homework Statement

I have a random, uniformly distributed vector with Cartesian components x,y,z. I should calculate the expectation value of the products of the components, e.g. $$<x\cdot x>, <x\cdot y>, ..., <z\cdot z>$$.

## Homework Equations

In spherical coordinates the $$x,y,z$$ components are

$$x = \sin(\theta)\cdot\cos(\phi)$$
$$y= \sin(\theta)\cdot\sin(\phi)$$
$$z = \cos(\theta)$$

$$<x\cdot x> =\int f(x)\cdot x^2 dx = const.\times \int x^2 dx$$,
because the probability density function is constant of the uniform probability distribution.

## The Attempt at a Solution

I think it is useful to convert the Cartesian coordinates to spherical coordinates. So for example

$$<x\cdot x> = \int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi~~[\sin(\theta)\cdot\cos(\phi)]^{2}$$

I wrote a program for this, and the solution is:

$$<x\cdot x> = <y\cdot y> = <z\cdot z> = \frac{1}{3}$$
and the other terms are zero.

But the solution of the above integral is not 1/3.

You don't need to convert to spherical coordinates, the limit you're assuming is from 0 to 1 on all three axes. And the value of the integral IS 1/3, but you've forgotten to multiply by f(x).

I don't know if youre calculating the same expectation value, but in QM, $$<f>=\int_{a}^{b}f(r)*f'*f(r)'dr$$,

where f' is the operator, f(r) is the function (physical quantity), and f(r)' is its conjugate. If f is real only, then f(r)=f(r)' and the integral becomes,

$$f(r)^2\int_{a}^{b}f'dr$$

If that is what you were trying to calculate, then you missed out by a factor of $$f(x)^2$$. Also, $$\theta$$ is a constant, not a variable.

It appears that you have made a mistake in your integration or in your understanding of the problem. The expectation value of a quantity is defined as the average value of that quantity over all possible outcomes, weighted by their probabilities. In this case, since the vector components are uniformly distributed, each component has an equal probability of being any value within its allowed range. Therefore, the expectation value of <x\cdot x> (or any other component product) should be the average value of that product over all possible values of x.

To calculate this, you can use the integral you have written, but with the probability density function (PDF) included:

<x\cdot x> = \int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi~~[\sin(\theta)\cdot\cos(\phi)]^{2} \cdot f(x) = \frac{1}{4\pi} \int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi~~[\sin(\theta)\cdot\cos(\phi)]^{2}

where the PDF, f(x), is equal to 1/(4\pi) for a uniform distribution over the surface of a sphere. Evaluating this integral gives the correct result of 1/3.

In general, when calculating expectation values, it is important to include the appropriate PDF in the integration to account for the weighting of different outcomes. Additionally, it may be helpful to check your work using a different method or by considering a simpler case where the answer is known.

## What is an expectation value?

An expectation value is a mathematical concept used in statistics and probability theory to represent the average or expected outcome of a random variable. It is calculated by multiplying each possible outcome by its probability and summing them together.

## Why is the expectation value important?

The expectation value is important because it provides a way to quantify the average outcome of a random variable and make predictions based on this value. It is also used in many statistical and physical models to represent the most likely outcome of a system.

## How is the expectation value calculated?

The expectation value is calculated by taking the product of each possible outcome of a random variable with its probability and summing them together. In mathematical notation, it can be represented as E(X) = ∑x P(x), where X is the random variable and P(x) is the probability of each outcome x.

## What factors can affect the expectation value?

The expectation value can be affected by various factors, such as the range and distribution of possible outcomes, the probabilities assigned to each outcome, and any external influences on the system being studied. It can also change if the underlying assumptions or conditions of the system are altered.

## How is the expectation value used in real-world applications?

The expectation value is used in a variety of real-world applications, such as finance, engineering, and physics. In finance, it is used to calculate the expected return on an investment. In engineering, it is used to predict the failure rate of a system. In physics, it is used to calculate the average energy of a particle in a given state.

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