1.5 Joint distributions

Let \(X_1, X_2, \dots , X_n\) be a random sample from the exponential distribution with mean \(1/\lambda \), and make the transformation \begin {gather*} Y_1 = X_1 , \quad Y_2 = X_1 + X_2 , \quad \dots , \\ Y_j = X_1 + \dots + X_j , \quad \dots , \quad Y_n = X_1 + \dots + X_n , \end {gather*} the successive partial sums. Clearly \(0 < Y_1 < Y_2 < \dots < Y_{n-1} < Y_n\). We want the joint pdf of \(Y_1, \dots , Y_n\), and from it the distribution of \(Y_n\).

The joint pdf of the sample is \[ f_{X_1,\dots ,X_n}(x_1, \dots , x_n) = \prod _{i=1}^n\lambda e^{-\lambda x_i} = \lambda ^ne^{-\lambda \sum _{i=1}^nx_i} . \] The inverse transformation is \[ X_1 = Y_1 , \quad X_2 = Y_2 - Y_1 , \quad X_3 = Y_3 - Y_2 , \quad \dots , \quad X_n = Y_n - Y_{n-1} , \] so the Jacobian matrix \(\bigl (\partial X_i/\partial Y_j\bigr )\) has \(1\) on the diagonal, \(-1\) just below it and \(0\) elsewhere: \[ J = \begin {vmatrix} 1 & 0 & 0 & \cdots & 0 \\ -1 & 1 & 0 & \cdots & 0 \\ 0 & -1 & 1 & \cdots & 0 \\ \vdots & & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & -1 & 1 \end {vmatrix} = 1 , \] the determinant of a triangular matrix being the product of its diagonal entries. Since \(\sum _ix_i = y_n\), \[ f_{Y_1,\dots ,Y_n}(y_1, \dots , y_n) = \lambda ^ne^{-\lambda y_n} , \qquad 0 < y_1 < y_2 < \dots < y_n . \] To find the density of \(Y_n\), integrate out \(y_1, y_2, \dots , y_{n-1}\) in turn, each over \(0 < y_j < y_{j+1}\): \[ \int _0^{y_2}dy_1 = y_2 , \qquad \int _0^{y_3}y_2\,dy_2 = \frac {y_3^2}{2} , \qquad \int _0^{y_4}\frac {y_3^2}{2}\,dy_3 = \frac {y_4^3}{2 \times 3} , \qquad \dots , \] and after \(n - 1\) integrations, \[ f_{Y_n}(y_n) = \lambda ^ne^{-\lambda y_n}\,\frac {y_n^{n-1}}{2 \times 3 \times \dots \times (n - 1)} = \frac {\lambda ^ny_n^{n-1}e^{-\lambda y_n}}{\Gamma (n)} , \qquad y_n > 0 , \] since \(\Gamma (n) = (n - 1)!\). This is the \(\text {Gamma}(n, \lambda )\) density. So, since each \(X_i \sim \text {Gamma}(1, \lambda )\), \[ Y_n = X_1 + X_2 + \dots + X_n \sim \text {Gamma}(n, \lambda ) . \]

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