4
$\begingroup$

Given that time interval $T^*$ in seconds between certain events has a negative exponential distribution.

The instrument cannot detect intervals which are less than $\delta$ seconds.

Let $T_1, ..., T_n$ be a sample of independent intervals measyred by the instrument. The distribution of one of those observation $T_i$ is the conditional distribution of $T^*$ given that $T^*>\delta$

In this question, if I want to find the probability density function of $T_i$, should I consider the shifted exponential distribution such that: $$f_T(t) = \begin{cases} \lambda e^{-\lambda (t- \delta)} & t>\delta, \\ 0 & otherwise \end{cases}$$

with $E(T)=\delta + \frac{1}{\lambda}$ and $Var(T) = \frac{1}{\lambda ^2}$

Thank you

$\endgroup$
  • $\begingroup$ From the memoryless property of the exponential distribution, $T^*-\delta|T^*>\delta$ has the same distribution as $T^*$. $\endgroup$ – Math1000 May 4 '15 at 0:31
  • $\begingroup$ Im wondering why the memoryless property can be applied to this case. The pdf depends on $T_i$ (not $T^*$), doesn't it? @Math1000 $\endgroup$ – Alana May 4 '15 at 0:49
  • $\begingroup$ I'm not sure I understand the relationship being asserted between the $T_i$ and $T^*$. Are you saying that there is a random variable, $T^*$, that represents an exponentially distributed interval of time in an experiment, and the $T_i$ are the measured values of $T^*$? And, furthermore, that $T_i = T^*$ if $T^* > \delta$, and $T_i$ is undefined otherwise? $\endgroup$ – Brian Tung May 4 '15 at 1:29
  • $\begingroup$ Also, by the way, for $t > \delta$, we should have $f_T(t) = \lambda e^{-\lambda(t-\delta)}$; the coefficient must be positive, else the PDF is negative. The "negative" of "negative exponential distribution" (usually just "exponential distribution") refers to the rate, not the coefficient. $\endgroup$ – Brian Tung May 4 '15 at 1:30
  • $\begingroup$ ye, thats what I mean, I edited the function :) thanks $\endgroup$ – Alana May 4 '15 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.