Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y:

` $$ \begin{aligned} f(x,y)=\left\{ \begin{array}{ll} xe^{-x(1+y)} & \hbox{$x\geq 0, y\geq 0$;}\\ 0, & \hbox{Otherwise} \end{array} \right. \end{aligned} $$ `

a. What is the probability that the lifetime X of the first component exceeds 3?

b. What are the marginal pdf's of X and Y? Are the two lifetimes independent? Explain.

c. What is the probability that the lifetime of at least one component exceeds 3?

### Solution

a. The probability that $X>3$ is found using the definition of the Joint Probability Density Function. As probability does not depend on the y values, we consider all possible values of y. So

` $$ \begin{aligned} P(X>3) &= \int_3^{\infty}\int_0^{\infty} f(x,y) \; dy dx\\ & = \int_3^{\infty}\int_0^{\infty} xe^{-x(1+y)} \; dy dx\\ & = \int_3^{\infty}\bigg(\int_0^{\infty} xe^{-x(1+y)} \; dy\bigg) dx\\ & = \int_3^{\infty}\bigg(\int_0^{\infty} xe^{-x} e^{-xy} \; dy\bigg) dx\\ & = \int_3^{\infty}xe^{-x} \bigg(\int_0^{\infty} e^{-xy} \; dy\bigg) dx\\ & = \int_3^{\infty}xe^{-x} \bigg[ \frac{e^{-xy}}{-x}\bigg]_0^{\infty}dx\\ & = \int_3^{\infty}e^{-x}dx\\ &= \bigg[-e^{-x}\bigg]_3^\infty \\ &= 0.05 \end{aligned} $$ `

b. In order to compute the marginal density function of X, we must integrate the function with respect to y.

The marginal density of $X$ is

` $$ \begin{aligned} f(x) &= \int_0^{\infty}f(x,y) \; dy\\ &= \int_0^{\infty} xe^{-x(1+y)} \; dy\\ &=\int_0^{\infty} xe^{-x} e^{-xy} \; dy\\ &=xe^{-x} \bigg(\int_0^{\infty} e^{-xy} \; dy\bigg)\\ &=xe^{-x}\bigg[ \frac{e^{-xy}}{-x}\bigg]_0^{\infty}\\ &= e^{-x}, x\geq 0 \end{aligned} $$ `

The marginal density of $Y$ is

` $$ \begin{aligned} f(y) &= \int_0^{\infty}f(x,y) \; dx\\ &= \int_0^{\infty} xe^{-x(1+y)} \; dx\\ &=\int_0^{\infty} xe^{-x} e^{-xy} \; dx\\ &=\frac{xe^{-x}}{-(1+y)}- \int \frac{xe^{-x}}{-(1+y)}dx\\ & = \frac{xe^{-x}}{-(1+y)}- \frac{xe^{-x}}{-(1+y)^2}\bigg|_0^\infty\\ &=0 - 0 - \bigg[0- \frac{1}{(1+y)^2} \bigg]\\ &= \frac{1}{(1+y)^2}, y\geq 0. \end{aligned} $$ `

Clearly, as $f(x,y)\neq f(x)\cdot f(y)$, $X$ and $Y$ are not independent.

c. The probability that the lifetime of at least one component exceeds 3 is

` $$ \begin{aligned} P[(X > 3)\cup (Y > 3)\cup (X > 3)\cap (Y > 3)] &= 1- P(X \leq 3\cap Y \leq 3)\\ &= 1- \int_0^3 \int_0^3 xe^{-x(1+y)} \; dy dx\\ &= 1- \int_0^3 \int_0^3 xe^{-x} e^{-xy} \; dy dx\\ &= 1- \int_0^3 e^{-x} (1-e^{-3x})\; dx\\ &= e^{-3} +0.25-0.25 e^{-12}\\ &=0.300. \end{aligned} $$ `

There is a 30% chance that at least one of the components lifetimes will exceed 3.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators