Question

Suppose \(X\) and \(Y\) have joint \(\mathrm{PDF}\) $$ f(x, y)=\left\\{\begin{array}{ll} e^{-x-y}, & \text { if } x \geq 0, y \geq 0 \\ 0, & \text { otherwise } \end{array}\right. $$ Find (a) the joint PDF of \(U=X+Y\) and \(V=X\) (b) the marginal PDF of \(U\).

Short answer

(a) Joint PDF: \( f_{U,V}(u,v) = e^{-u} \) for \( U \geq V \geq 0 \). (b) Marginal PDF: \( f_U(u) = u e^{-u} \) for \( u \geq 0 \).

Step-by-step solution

Determine the Range for U and V

First, identify that since \(X \geq 0\) and \(Y \geq 0\), the variable \(U = X + Y\) must be \(U \geq 0\). Also, \(V = X\) implies \(V \geq 0\). Therefore, the transformation is valid for \(U \geq V \geq 0\).

Transform the Variables

We use the transformation \(U = X + Y\) and \(V = X\). Since \(Y = U - V\), we have that \(Y \geq 0\) implies \(U \geq V\). Therefore, the support of \( (U, V) \) is \( U \geq V \geq 0\).

Calculate the Jacobian of the Transformation

The transformation from \((X,Y)\) to \((U,V)\) requires us to compute the Jacobian. The transformation functions are:\(U = X + Y\), \(V = X\), so\[\begin{vmatrix}\frac{\partial U}{\partial X} & \frac{\partial U}{\partial Y}\\frac{\partial V}{\partial X} & \frac{\partial V}{\partial Y}\end{vmatrix} = \begin{vmatrix}1 & 1 \1 & 0\end{vmatrix} = -1\]Thus, the Jacobian \(J = | -1 | = 1\).

Write the Joint PDF of U and V

Since the joint PDF of \(X\) and \(Y\) is \(f(x,y) = e^{-x-y}\) for \(x \geq 0\) and \(y \geq 0\), the joint PDF of \(U\) and \(V\) becomes:\[f_{U,V}(u,v) = f(v, u-v) \times J = e^{-v - (u-v)} = e^{-u}\]The support is \(U \geq V \geq 0\).

Find the Marginal PDF of U

To find the marginal PDF of \(U\), integrate the joint PDF over \(V\):\[f_U(u) = \int_0^u f_{U,V}(u,v) \, dv = \int_0^u e^{-u} \, dv\]Since \(e^{-u}\) is constant with respect to \(v\), this becomes:\[f_U(u) = e^{-u}(u - 0) = ue^{-u}\]for \(u \geq 0\).

Simplify the Marginal PDF of U

The marginal PDF derived simplifies to showing that:- \(f_U(u) = u e^{-u}\), \(u \geq 0\).- It is clear that the PDF is zero for \(u < 0\).

Key concepts

Transformation of Variables

In probability theory, transforming variables is essential when you need to understand how a new set of variables relates to an original set. Here, we have original variables \(X\) and \(Y\), and we're transforming them into \(U\) and \(V\). The transformation given is \(U = X + Y\) and \(V = X\). This means \(Y\) can be expressed as \(U - V\).
For a transformation to work, it's crucial to delineate the range over which the transformation is valid. Since \(X \geq 0\) and \(Y \geq 0\), it follows that \(U = X + Y \geq 0\). Similarly, \(V = X \geq 0\). Also, since \(Y = U - V \geq 0\), the condition \(U \geq V\) must be satisfied. These conditions provide necessary constraints for the new variables \(U\) and \(V\).
During variable transformations, it's important to express all relevant conditions and check that they cover all scenarios within the region of interest for the new variables.

Jacobian Determinant

The Jacobian determinant is a fundamental part of transforming variables when dealing with probability density functions. It essentially adjusts the scales and transforms the density accordingly. For the given transformation from \((X, Y)\) to \((U, V)\), the Jacobian helps us relate these two sets of variables accurately.
The transformation functions are \( U = X + Y \) and \( V = X \). Therefore, we compute the Jacobian determinant as follows:
\[\begin{vmatrix}\frac{\partial U}{\partial X} & \frac{\partial U}{\partial Y} \\frac{\partial V}{\partial X} & \frac{\partial V}{\partial Y}\end{vmatrix} = \begin{vmatrix}1 & 1 \ 1 & 0\end{vmatrix} = -1\]
The absolute value of this determinant, \(| -1 | = 1\), is used in transforming the joint probability density function into the new variables \(U\) and \(V\).
Remember, the Jacobian determinant tells us how volume scales when moving from one variable space to another. In our context, since the Jacobian is \(1\), there is no scaling effect and thus, density values remain unaffected after transformation.

Marginal Probability Density Function

The marginal probability density function (PDF) helps us find the probability distribution of one variable irrespective of other variables involved. To retrieve the marginal PDF for \(U\) from the joint PDF of \(U\) and \(V\), we integrate out the variable \(V\).
Here, the joint PDF \(f_{U,V}(u,v)\) is defined as \(e^{-u}\) for \(U \geq V \geq 0\). To find the marginal PDF \(f_U(u)\), integrate over the possible values of \(V\):
\[f_U(u) = \int_0^u e^{-u} \, dv\]
Because \(e^{-u}\) is constant with respect to \(v\), the integration simplifies to:
\[f_U(u) = e^{-u} \int_0^u \, dv = e^{-u}(u - 0) = ue^{-u}\]
Thus, \(f_U(u)\) represents the likelihood of the value \(U\) taking within its domain. This marginal PDF is only valid for \(u \geq 0\), with the function evaluating to zero for \(u < 0\). This step emphasizes the importance of integration in probability for marginalizing variables and simplifying conditional relationships.

Integration in Probability

Integration plays a critical role in solving probability problems, especially when determining marginal distributions and normalizing probability density functions. When dealing with joint PDFs, integrating over one variable allows us to focus on the other.
In the context of our exercise, integration helps extract the marginal PDF of \(U\) from the joint PDF \(f_{U,V}(u,v)\). It involves calculating:
\[f_U(u) = \int_0^u e^{-u} \, dv\]
This integral simplifies because our function, \(e^{-u}\), is constant with respect to \(v\), turning the problem essentially into multiplying by the interval length from \(0\) to \(u\).
Integration in probability also serves to ensure total probability remains unified (i.e., the total area under the curve is 1) and aids in finding expected values and variances.
Always verify that integration respects the function's domain and constraints, ensuring probabilistic interpretations stay intact throughout the problem-solving process.