3 Exercício 2

Considere que X \(\sim\) exp(\(\lambda\)). Use o teorema da convolução para determinar a função de densidade de probabilidade de Y = \(X_1\) + \(X_2\) + \(X_3\) + \(X_4\) sobre a condição de indepêndencia.

\[\begin{equation} \begin{split} Y_1 = X_1 \sim exp(\lambda); x \geq 0 &\\ X_2 \sim exp(\lambda); x \geq 0 &\\ X_3 \sim exp(\lambda); x \geq 0 &\\ X_4 \sim exp(\lambda); x \geq 0 & \end{split} \begin{split} & = f(x) \left\{\begin{matrix} \lambda e^{\lambda x} & ; x\geq0 \\ 0 & CC \end{matrix}\right. \end{split} \end{equation}\]

Usando o teorema da convolução

\[\begin{equation} \begin{split} Y_2 &= \int_{-\infty}^{\infty}f_{Y_1}(y-x_2)f_{X_2}(x_2)dx_2 \\ &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(y - x_2)} \lambda e^{-x_2} dx_2 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda y}dx_2 \end{split} \end{equation}\]

Como \(x_2 \geq 0\) e \(y - x_2 \geq 0\) temos então que \(0 \leq x_2 \leq y\).

\[\begin{equation} \begin{split} Y_2 &= \lambda^2 e^{-\lambda y} \int_{0}^{y}dx_2 \\ &= \lambda^2 e^{-\lambda y } y \ \ ; \ \ y \geq 0 \end{split} \end{equation}\]

Portanto temos que:

\[Y_2 = \left\{\begin{matrix} \lambda^2 y e^{-\lambda y} & y \geq 0 \\ 0 & cc \end{matrix}\right. \]

Para \(Y_3\).

\[\begin{equation} \begin{split} Y_3 & = \int_{-\infty}^{\infty} fy_2(y - x_3) fx_3(x_3)dx_3 \\ & = \int_{-\infty}^{\infty} \lambda^2 (y - x_3) e^{-\lambda (y - x_3)} \lambda e^{-\lambda x_3}dx_3\\ & = \int_{-\infty}^{\infty} \lambda^3 e^{- \lambda y} (y-x_3)dx_3 \end{split} \end{equation}\]

Como \(x_3 \geq 0\) e \(y- x_3 \geq 0\) temos então que \(0 \leq x_3 \leq y\).

\[\begin{equation} \begin{split} Y_3 & = \int_{0}^{y} \lambda^3 e^{- \lambda y}y dx_3 - \int_{0}^{y} \lambda^3 e^{- \lambda y} x_3 dx_3\\ & = \lambda^3 e^{- \lambda y} y \int_{0}^{y} dx_3 - \lambda^3 e^{- \lambda y} \int_{0}^{y} x_3 dx_3\\ & = \lambda^3 e^{- \lambda y} y y - \lambda^3 \frac{y^2}{2} e^{- \lambda y}\\ & = \lambda^3 e^{- \lambda y} y^2 - \lambda^3 \frac{y^2}{2} e^{- \lambda y} \\ & = \lambda^3 \frac{y^2}{2} e^{- \lambda y} ; y \geq 0 \end{split} \end{equation}\]

Portanto temos que:

\[Y_3 = \left\{\begin{matrix} \lambda^3 \frac{y^2}{2} e^{- \lambda y} & y \geq 0 \\ 0 & cc \end{matrix}\right. \]

Para \(Y_4\).

\[\begin{equation} \begin{split} Y_4 & = \int_{-\infty}^{\infty} fy_3(y - x_4) fx_4(x_4)dx_4 \\ & = \int_{-\infty}^{\infty} \lambda^3 \frac{(y - x_4)^2}{2} e^{- \lambda (y - x_4)} \lambda e^{-\lambda x_4}dx_4\\ & = \int_{-\infty}^{\infty} \lambda^4 \frac{(y - x_4)^2}{2} e^{- \lambda y}dx_4\\ \end{split} \end{equation}\]

Como \(x_4 \geq 0\) e \(y- x_4 \geq 0\) temos então que \(0 \leq x_4 \leq y\).

\[\begin{equation} \begin{split} Y_4 & = \int_{0}^{y} \lambda^4 \frac{(y^2 -2yx_4 + x_4^2)}{2} e^{-\lambda y}dx_4\\ & = \int_{0}^{y} \lambda^4 e^{-\lambda y} \frac{y^2}{2} dx_4 - 2 \int_{0}^{y} \lambda^4 e^{-\lambda y} \frac{-yx_4}{2} dx_4 + \int_{0}^{y} \lambda^4 e^{-\lambda y} \frac{x_4^2}{2} dx_4\\ & = \lambda^4 e^{-\lambda y} \frac{y^2}{2} \int_{0}^{y} dx_4 - \lambda^4 e^{-\lambda y}y \int_{0}^{y} x_4 dx_4 + \lambda^4 e^{-\lambda y} \int_{0}^{y} \frac{x_4^2}{2}dx_4 \\ & = \lambda^4 e^{-\lambda y}\frac{y^3}{2} - \lambda^4 e^{-\lambda y} \frac{y^3}{2} + \lambda^4 e^{-\lambda y} \frac{y^3}{6} \\ & = \lambda^4 e^{-\lambda y} \frac{y^3}{6}; y \geq 0 \end{split} \end{equation}\]

Portanto temos que:

\[Y_4 = \left\{\begin{matrix} \lambda^4 \frac{y^3}{6} e^{-\lambda y} & y \geq 0 \\ 0 & cc \end{matrix}\right. \]