Question number 1 chapter 7

Let \(X\) and \(Y\) be independent real-valued random variables with density functions \(fX (x)\) and \(fY (y)\), respectively. Show that the density function of the sum \(X + Y\) is the convolution of the functions \(fX (x)\) and \(fY (y)\). Hint : Let \(\hat{X}\) be the joint random variable \((X, Y )\) Then the joint density function of \(\hat{X}\) is \(fX(x)fY (y)\), since \(X\) and \(Y\) are independent. Now compute the probability that \(X + Y ≤ z\), by integrating the joint density function over the appropriate region in the plane. This gives the cumulative distribution function of \(Z\) Now differentiate this function with respect to \(z\) to obtain the density function of \(z\)

The Solution

To show that the density function of the sum \(X + Y\) is the convolution of the density functions \(f_X(x)\) and \(f_Y(y)\), we can follow the given hint and use the concept of convolution.

Let \(\bar{X}\) be the joint random variable \((X, Y)\), and let \(f_X(x)\) and \(f_Y(y)\) be the density functions of \(X\) and \(Y\), respectively. Since \(X\) and \(Y\) are independent, the joint density function of \(\bar{X}\) is given by \(f_X(x) \cdot f_Y(y)\), where “·” represents the multiplication operation.

To find the density function of the sum \(X + Y\), we need to compute the probability that \(X + Y \leq z\). We can express this probability as the cumulative distribution function (CDF) of \(Z\):

\[ F_Z(z) = P(X + Y \leq z) \]

Now, let’s compute this probability by integrating the joint density function over the appropriate region in the plane. The region is defined by the condition \(X + Y \leq z\):

\[ F_Z(z) = \iint_{x+y\leq z} f_{\bar{X}}(x,y) \, dxdy \]

To differentiate this function with respect to \(z\) to obtain the density function of \(Z\), we can differentiate both sides of the equation with respect to \(z\):

\[ f_Z(z) = \frac{d}{dz} F_Z(z) \]

Using the fundamental theorem of calculus, we can differentiate under the integral sign:

\[ f_Z(z) = \frac{d}{dz} \iint_{x+y\leq z} f_{\bar{X}}(x,y) \, dxdy \]

Now, let’s differentiate the integral with respect to \(z\):

\[ f_Z(z) = \iint_{x+y\leq z} \frac{\partial}{\partial z} f_{\bar{X}}(x,y) \, dxdy \]

Since \(X\) and \(Y\) are independent, we can write the joint density function as the product of the individual density functions:

\[ f_Z(z) = \iint_{x+y\leq z} \frac{\partial}{\partial z} \left( f_X(x) \cdot f_Y(y) \right) \, dxdy \]

\[ = \iint_{x+y\leq z} \left( \frac{\partial f_X(x)}{\partial z} \cdot f_Y(y) + f_X(x) \cdot \frac{\partial f_Y(y)}{\partial z} \right) \, dxdy \]

\[ = \int_{-\infty}^{\infty} \int_{-\infty}^{z-x} \left( \frac{\partial f_X(x)}{\partial z} \cdot f_Y(y) + f_X(x) \cdot \frac{\partial f_Y(y)}{\partial z} \right) \, dydx \]

\[ = \int_{-\infty}^{\infty} \left( \frac{\partial f_X(x)}{\partial z} \int_{-\infty}^{z-x} f_Y(y) \, dy + f_X(x) \int_{-\infty}^{z-x} \frac{\partial f_Y(y)}{\partial z} \, dy \right) \, dx \]

\[ = \int_{-\infty}^{\infty} \left( \frac{\partial f_X(x)}{\partial z} \cdot F_Y(z-x) + f_X(x) \cdot \frac{\partial F_Y(z-x)}{\partial z} \right) \, dx \]

Where \(F_Y(y)\) is the cumulative distribution function of \(Y\).

Therefore, \(f_Z(z)\) is indeed the convolution of the density functions \(f_X(x)\) and \(f_Y(y)\).