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Suppose \(X\) and \(Y\) are independent continuous random variables. Show that $$ E[X \mid Y=y]=E[X] \text { for all } y $$

Short Answer

Expert verified
Using the definition of conditional expectation and the property of independence between random variables X and Y, we can find the conditional probability density function of X given Y = y. We then substitute this back into the definition of conditional expectation and recognize that the integral in the expression is the expected value of X. Thus, for independent continuous random variables X and Y, the expected value of X given Y equals y is the same as the expected value of X: \(E[X \mid Y=y] = E[X]\). This holds for all y.

Step by step solution

01

Write down the definition of conditional expectation

To find the conditional expectation of the random variable X given Y equals y, we can use the definition: \(E[X \mid Y=y] = \int_{-\infty}^{\infty} x \cdot f_{X \mid Y}(x \mid y) dx\) where \(f_{X \mid Y}(x \mid y)\) is the conditional probability density function of X given Y = y.
02

Use the property of independence

Since X and Y are independent random variables, their joint probability density function can be expressed as the product of their marginal probability density functions: \(f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)\) As a consequence, the conditional probability density function of X given Y = y is: \(f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{f_X(x) \cdot f_Y(y)}{f_Y(y)}\) Since \(f_Y(y) \neq 0\), we can simplify the expression: \(f_{X \mid Y}(x \mid y) = f_X(x)\)
03

Substitute the conditional PDF

Now that we have the conditional probability density function for independent random variables X and Y, we can substitute it back into the definition of the conditional expectation: \(E[X \mid Y=y] = \int_{-\infty}^{\infty} x \cdot f_{X \mid Y}(x \mid y) dx = \int_{-\infty}^{\infty} x \cdot f_X(x) dx\)
04

Recognize the expected value of X

The integral in the last expression represents the expected value of the random variable X: \(E[X] = \int_{-\infty}^{\infty} x \cdot f_X(x) dx\)
05

Make the conclusion

Since the integral in Step 3 is equal to the expected value of X, we can conclude that: \(E[X \mid Y=y] = E[X]\) This holds for all y, as required. Therefore, the expected value of X given Y equals y is the same as the expected value of X for independent random variables X and Y.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Random Variables
When we talk about random variables, we refer to measurable and quantifiable factors that can take different values based on the outcomes of some random phenomena. Now, consider two such variables, called independent random variables. The term "independent" in probability and statistics indicates that the outcome of one variable does not affect the outcome of the other.
For example, let's say we have two independent random variables, X and Y. If X represents the roll of a dice and Y represents the toss of a coin, the result of the dice roll will not influence the coin toss. Essentially, knowing the value of Y does not give us any information about X and vice versa.
In mathematical terms, the joint probability density function of two independent random variables can be expressed as the product of their marginal probability density functions:

  • For independent X and Y: \( f_{X,Y}(x,y) = f_X(x) \times f_Y(y) \)

This property is foundational in simplifying many problems involving probability calculations with multiple variables.
Probability Density Function
The probability density function (PDF) is a crucial concept when discussing continuous random variables. A PDF describes the likelihood of a random variable to take on a particular value. For continuous random variables, unlike discrete random variables, we do not get a probability for a specific value. Instead, we calculate the probability that the variable will fall within a certain range.
The PDF, denoted as \(f(x)\), provides a way to visualize the distribution of the random variable's possible values. Its graph is often a curve, and the area under the curve between two points gives the probability that the random variable falls between those two points.
Importantly, the PDF must satisfy two properties:
  • It must be non-negative, meaning \(f(x) \geq 0 \) for all \(x\).
  • The total integral of the PDF over all its possible values must equal 1, ensuring a full distribution of outcomes.

    \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \]
Marginal Probability Density Function
Marginal probability density functions are derived from joint probability functions. When dealing with multiple random variables, the joint PDF explains the probability distribution over all variables. However, sometimes we are interested in the probability distribution of just one of these variables, which leads us to the concept of the marginal PDF.
The marginal PDF of a particular variable, say \( X \), is found by integrating the joint PDF over the other variable(s). Let's consider random variables X and Y with the joint PDF \( f_{X,Y}(x,y) \). To get the marginal PDF of X, denoted as \( f_X(x) \), we integrate over all possible values of Y:

\[ f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy \]
Similarly, to find the marginal PDF of Y, we integrate the joint PDF over X. This concept simplifies problems by reducing the complexity of multi-variable probability distributions to just one variable.
Expected Value
The expected value, also called the mean, of a random variable is one of the fundamental notions in probability. It provides the average or "long-run" expected outcome of random variables over numerous trials. The expected value is like the center of the distribution of a random variable.
For a continuous random variable X with a probability density function \( f(x) \), the expected value \( E[X] \) is defined by the integral:

\[ E[X] = \int_{-\infty}^{\infty} x \, f(x) \, dx \]
This integral sums up all possible values of X, each weighted by its probability, to provide the average outcome we expect if we observe the random variable many times.
The concept of expected value is generalizable to more complex probability scenarios. For independent random variables like in the exercise, we find that conditional expectations simplify due to the independence property, leading to results like \( E[X \mid Y=y] = E[X] \), further illustrating the robust and insightful nature of expected values in probability.

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