Question

Prove that if \(X\) and \(Y\) are jointly continuous, then $$ E[X]=\int_{-\infty}^{\infty} E[X \mid Y=y] f_{Y}(y) d y $$

Short answer

To prove the given formula, first calculate the marginal probability density function of \(Y\), \(f_Y(y)\), by integrating the joint probability density function of \(X\) and \(Y\), \(f_{X,Y}(x,y)\), with respect to \(x\). Next, express the conditional probability density function of \(X\) given \(Y = y\), \(f_{X \mid Y}(x \mid y)\), using the joint probability density function and the marginal probability density function of \(Y\). Then, calculate the conditional expectation of \(X\) given \(Y = y\), \(E[X \mid Y=y]\), by integrating the product of \(x\) and the conditional probability density function of \(X\) given \(Y = y\). Finally, compute the expected value of \(X\), \(E[X]\), by integrating the product of the conditional expectation and the marginal probability density function of \(Y\). The resulting expression is \(E[X] = \int_{-\infty}^{\infty} E[X \mid Y=y] f_{Y}(y) dy\).

Step-by-step solution

Calculate the marginal probability density function of Y

We will begin by calculating the marginal probability density function \(f_Y(y)\). This is found by integrating the joint probability density function with respect to \(x\), i.e., \(f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x, y) dx \).

Express the conditional probability density function of X given Y = y

Next, we express the conditional probability density function \(f_{X \mid Y}(x \mid y)\) using the joint probability density function and the marginal probability density function of \(Y\). This is given by the following formula: \(f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}\).

Calculate the conditional expectation of X given Y = y

Now we will calculate the conditional expectation of \(X\) given \(Y = y\). This is computed by integrating the product of \(x\) and the conditional probability density function of \(X\) given \(Y = y\). Formally, this can be written as: \(E[X \mid Y=y] = \int_{-\infty}^{\infty} x \cdot f_{X \mid Y}(x \mid y) dx\).

Compute the expected value of X

Finally, we will compute the expected value of \(X\), by integrating the product of the conditional expectation of \(X\) given \(Y = y\) and the marginal probability density function of \(Y\). In other words, we will prove that: \(E[X] = \int_{-\infty}^{\infty} E[X \mid Y=y] f_{Y}(y) dy\). We will need to substitute the expression for the conditional expectation and marginal probability density function that we derived in the previous steps. \(E[X] = \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} x \cdot \frac{f_{X,Y}(x,y)}{f_Y(y)} dx \right) \left( \int_{-\infty}^{\infty} f_{X,Y}(x, y) dx \right) dy\). Now, notice that we can simplify the above expression and cancel out the marginal probability density function of \(Y\). \(E[X] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x \cdot f_{X,Y}(x,y) dx dy\). This is the expected value of \(X\), calculated by integrating the product of \(x\) and the joint probability density function of \(X\) and \(Y\). Thus, we have successfully proved the given formula, \(E[X] = \int_{-\infty}^{\infty} E[X \mid Y=y] f_{Y}(y) dy\).