Answer to Question #136740 in Statistics and Probability for sam markson

Question

. Suppose that the time in days until hospital discharge for a certain patient population follows a density f(x) = (1/3.3)exp(−x/3.3) for x > 0.
a. Find the mean and variance of this distribution.
b. The general form of this density (the exponential density) is f(x) =(1/β)exp(−x/β) for x > 0 for a fixed value of β. Calculate the mean and variance of this density.

Answer 1

$$f(x)=\dfrac{e^{-x/3.3}}{3.3}, x>0$$

$E(X)=\displaystyle\int_{-\infin}^\infin xf(x)dx=\displaystyle\int_{0}^\infin x\dfrac{e^{-x/3.3}}{3.3}dx$

$$\int x\dfrac{e^{-x/3.3}}{3.3}dx=-xe^{-x/3.3}+\int e^{-x/3.3}dx=$$

$$=-xe^{-x/3.3}-3.3e^{-x/3.3}+C$$

$$E(X)=\lim\limits_{A\to \infin}\big[-xe^{-x/3.3}-3.3e^{-x/3.3}\big]\begin{matrix} A \\ 0 \end{matrix}=3.3$$

$$E(X^2)=\displaystyle\int_{-\infin}^\infin x^2f(x)dx=\displaystyle\int_{0}^\infin x^2\dfrac{e^{-x/3.3}}{3.3}dx$$

$$\int x^2\dfrac{e^{-x/3.3}}{3.3}dx=-x^2e^{-x/3.3}+2\int xe^{-x/3.3}dx=$$

$$=-x^2e^{-x/3.3}-6.6xe^{-x/3.3}-21.78e^{-x/3.3}+C$$

$$E(X^2)=\lim\limits_{A\to \infin}\big[-x^2e^{-x/3.3}-6.6xe^{-x/3.3}-21.78e^{-x/3.3}\big]\begin{matrix} A \\ 0 \end{matrix}=$$ $$=21.78$$

$V(X)=E(X^2)-(E(X))^2=21.78-(3.3)^2=10.89$

b.

$$f(x)=\dfrac{e^{-x/\beta}}{\beta}, x>0$$

$E(X)=\displaystyle\int_{-\infin}^\infin xf(x)dx=\displaystyle\int_{0}^\infin x\dfrac{e^{-x/\beta}}{\beta}dx$

$$\int x\dfrac{e^{-x/\beta}}{\beta}dx=-xe^{-x/\beta}+\int e^{-x/\beta}dx=$$ $$=-xe^{-x/\beta}-\beta e^{-x/\beta}+C$$

$$E(X)=\lim\limits_{A\to \infin}\big[-xe^{-x/\beta}-\beta e^{-x/\beta}\big]\begin{matrix} A \\ 0 \end{matrix}=\beta$$

$$E(X^2)=\displaystyle\int_{-\infin}^\infin x^2f(x)dx=\displaystyle\int_{0}^\infin x^2\dfrac{e^{-x/\beta}}{\beta}dx$$

$$\int x^2\dfrac{e^{-x/\beta}}{\beta}dx=-x^2e^{-x/\beta}+2\int xe^{-x/\beta}dx=$$

$$=-x^2e^{-x/\beta}-2\beta xe^{-x/\beta}-2\beta ^2e^{-x/\beta}+C$$

$$E(X^2)=\lim\limits_{A\to \infin}\big[-x^2e^{-x/\beta}-2\beta xe^{-x/\beta}-2\beta^2e^{-x/\beta}\big]\begin{matrix} A \\ 0 \end{matrix}=$$ $$=2\beta^2$$

$V(X)=E(X^2)-(E(X))^2=2\beta^2-(\beta)^2=\beta^2$

$E(X)=\beta$

$V(X)=\beta^2$