Question

Suppose that T is an Exponential (λ=1) random variable and f(x)≐⌊x⌋

i.e. is the largest integer not more than x.

Find the cdf and pmf for X≐f(T). What is E[f(T)]?

Answer 1

$PMF~of~X=[T]~is\\ P(X=x)=P([T]=x)=P(x\leq T<x&plus;1)\\ =\int_x^{x&plus;1}e^{-t}dt=e^{-x}-e^{-x-1}=e^{-x}(1-e^{-1});~x=0,1,2,3,...\\ =0~otherwise\\ CDF~of~X~is\\ F_X(x)=P(X\leq x)=0~if~x<0\\ =\sum_{i=0}^xe^{-i}(1-e^{-1})=(1-e^{-1})(1&plus;e^{-1}&plus;e^{-2}&plus;...&plus;e^{-x})=(1-e^{-1})\times\frac{1-e^{-x-1}}{1-e^{-1}}\\ =1-e^{-x-1}~if~x\geq 0\\$

$Take~e^{-1}=q,~say~then~1-q=p=1-e^{-1}\\ then~E(X)=\sum_{x=0}^\infty xpq^x=pq\sum_{x=0}^\infty \frac{\mathrm{d} q^x}{\mathrm{d} q}=pq\frac{\mathrm{d} }{\mathrm{d} q}\left(\sum_{x=0}^\infty q^x \right )=pq\times \frac{\mathrm{d} }{\mathrm{d} q}\left(\frac{1}{1-q} \right )\\ =pq\times \frac{1}{(1-q)^2}=\frac{q}{p}=\frac{e^{-1}}{1-e^{-1}}$