# Order Statistics Example : Electronic components of a certain type…

Electronic components of a certain type have a length life (in hours) X, that follows the exponential distribution with probability density given by $$f(x) = \frac{1}{100}e^{-x/100}, x > 0$$

a. Suppose that 2 such components operate independently and in series in a certain system (that is, the system fails when either component fails). Find the density function for the length of life of the system.

b. Suppose that 2 such components operate independently and in parallel in a certain system (that is, the system does not fail until both components fail). Find the density function for the length of life of the system.

a. Let the random variables $X_1, X_2$ denote the length of the life of the electronic components and let the random variable $M:=\min\{X_1,X_2\}$ which denotes the length of the life of the system (since it fails as soon as the first components fail, i.e. it's lifetime is equal to the minimum lifetime of the two components). We find firstly the probability distribution function $F_M$ of $M$. It is \begin{align*}F_M(x)&=P(M\leq x)=1-P(M\geq x)=1-P(\min\{X_1,X_2\}\geq x)=\\&=1-P(X_1\geq x, X_2 \geq x)=^{\text{independent}}\\&=1-P(X_1\geq x)\cdot P(X_2\geq x)=^{\text{equally distributed}}\\&=1-[P(X\geq x)]^2=1-[1-P(X\leq x)]^2=1-[1-F_X(x)]^2\end{align*} where $F_X$ denotes the probability distribution function of the given exponential prbabality density function. The desired probability density function $f_M$ is the derivative of $F_M$, that is $$f_M(x)= \frac{dF_M(x)}{dx}=\frac{d}{dx}(1-[1-F_X(x)]^2)=-2[1-F_X(x)](1-F_X(x))'=-2[1-F_X(x)](-f_X(x))=2f_X(x)[1-F_X(x)]$$ and by substituting the exponential formula for $f_X(x)=\frac{1}{100}e^{-\frac{x}{100}}$ and $F_X(x)=1-e^{-\frac{x}{100}}$ you have the result $$f_M(x)=2\frac{1}{100}e^{-\frac{x}{100}}[1-(1-e^{-\frac{x}{100}})]=\frac{1}{50}e^{-\frac{x}{50}}$$ i.e. the length of the life of the system has the exponential distribution with parameter $\frac{1}{50}$. That is a well known property,. i.e. that the minimum of two exponential random variables with parameters $\lambda_1$ and $\lambda_2$ is again exponentially distributed with parameter $\lambda_1+\lambda_2$ (here $\lambda_1=\lambda_2=\frac{1}{100}$).
b. Now we want to calculate the probability density function of the sum of the two random variables $X_1$ and $X_2$. It is again well known (see Wikipedia) that the sum of two independent identically distributed (i.e. with the same $\lambda$) exponential random variables has the Gamma (or Erlang) distribution with parameters $2$ and $\lambda$. For a proof with the use of moment generating functions see here http://www.statlect.com/ucdexp1.htm. or here http://ocw.mit.edu/courses/mathematics/18-440-probability-and-random-variables-spring-2011/lecture-notes/MIT18_440S11_Lecture23.pdf or you can find many other sources under the search "sum of exponential random variables is gamma proof".