Question-21
It is given that \(P( X\geqslant 2) =0.25\) for an exponentially distributed random variable \(X\) with \(E[ X] =\cfrac{1}{\lambda }\), where \(E[ X]\) denotes the expectation of \(X\). What is the value of \(\lambda\)? (\(\ln\)denotes natural logarithm)
The mean of an exponentially distributed random variable with parameter \(\lambda\) is \(\cfrac{1}{\lambda }\). The CDF is given by
\[ F_{X}( x) =\begin{cases} 1-e^{-\lambda x} , & x\geqslant 0\\ 0, & \text{otherwise} \end{cases} \] Now:
\[ \begin{array}{ r r l } & P( X\geqslant 2) & =0.25\\ \Longrightarrow & 1-P( X< 2) & =0.25\\ \Longrightarrow & P( X< 2) & =0.75\\ \Longrightarrow & F_{X}( 2) & =0.75\\ \Longrightarrow & 1-e^{-2\lambda } & =0.75\\ \Longrightarrow & e^{-2\lambda } & =0.25\\ \Longrightarrow & \lambda & =-\cfrac{1}{2}\ln 0.25\\ \Longrightarrow & \lambda & =\ln 2 \end{array} \]