Question-41
For \(x\in \mathbb{R}\), the floor function is denoted by \(f( x) =\lfloor x\rfloor\) and defined as follows
\[ \lfloor x\rfloor =k,k\leqslant x< k+1 \]
where \(k\) is an integer. Let \(Y=\lfloor X\rfloor\), where \(X\) is an exponentially distributed random variable with mean \(\cfrac{1}{\ln 10}\), where \(\ln\) denotes the natural logarithm. For any positive integer \(l\), one can write the probability of the event \(Y=l\) as follows:
\[ P( Y=l) =q^{l}( 1-q) \]
The value of \(q\) is
We have:
\[ \begin{aligned} P( Y=l) & =P( l\leqslant X< l+1)\\ & =F_{X}( l+1) -F_{X}( l) \end{aligned} \]
Recall that the CDF of the exponential distribution with parameter \(\lambda\) is given by:
\[ F_{X}( x) =\begin{cases} 1-e^{-\lambda x} , & x\geqslant 0\\ 0, & x< 0 \end{cases} \]
The mean of \(X\) is given by \(\cfrac{1}{\lambda }\).
\[ \begin{aligned} P( Y=l) & =\left( 1-e^{-\lambda ( l+1)}\right) -\left( 1-e^{-\lambda l}\right)\\ & =e^{-\lambda l} -e^{-\lambda l-\lambda }\\ & =e^{-\lambda l}\left[ 1-e^{-\lambda }\right]\\ & =\left( e^{-\lambda }\right)^{l}\left[ 1-\left( e^{-\lambda }\right)\right] \end{aligned} \]
We see that \(q=e^{-\lambda }\). In this case, we have \(\lambda =\ln 10\), which makes \(q=e^{-\lambda } =0.1\). Note how this is closely related to the geometric distribution.