The CDF for an exponential distribution of rate $\lambda$ truncated at T is

$F(t) = \frac{1-e^{-\lambda t}}{1-e^{-\lambda T}}$. (for $t<T$, else 0).

I would like to determine $\lambda$ and $T$ such that F(5) = 0.95, and F(2) = 0.8. (So 80% of events happen within time 2, and 95% of events happen within time 5). However, it would seem these equations have no solution. Is this actually true? I'm struggling to see why I can't find a truncated exponential satisfying these conditions? Have I made a mistake somewhere?

I entered the equations in Mathematica to find that there wasn't a solution. But similarly,

\begin{gather} F(5) = \frac{1-e^{-5\lambda}}{1-e^{-\lambda T}} \\ F(2) = \frac{1-e^{-2\lambda}}{1-e^{-\lambda T}}. \\ \end{gather}

Calculating the ratio of these, I get \begin{equation} \frac{F(5)}{F(2)} = \frac{1-e^{-5\lambda}}{1-e^{-2\lambda}} = \frac{19}{16} \end{equation}.

This can be solved for $\lambda$ which gives $\lambda \approx 0.89316$. Putting this back into the expression for $F(5)$, we end up wanting to solve

\begin{equation} F(5) = \frac{1-e^{-5\lambda}}{1-e^{-\lambda T}} \approx \frac{0.9885}{1-e^{-\lambda T}}. \end{equation}

However, the numerator is already greater than the value I want it to be, so there's now way that this equation can equal 0.95. Is my thinking flawed here? Are there some other roots that I am somehow missing? Or are the numerical errors caused by putting in rounded values (to 4-5dp) the cause of this?

  • $\begingroup$ The equations usually do have a solution. You will find them easier to solve by writing $\eta_i=1-F(t_i)(1-\exp(-\lambda T)),$ which leads to the system of linear equations $\lambda t_i=\eta_i.$ Solve that for $\lambda$ and then solve the original equations for $T.$ $\endgroup$ – whuber Apr 16 at 13:43
  • $\begingroup$ Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ – kjetil b halvorsen Apr 16 at 14:02
  • $\begingroup$ @kjetilbhalvorsen Apologies. I have added some more detail. For what it's worth, this isn't a homework question. It's a problem I came across trying to do something else. But I appreciate it does seem that way. $\endgroup$ – user112495 Apr 16 at 14:30
  • $\begingroup$ @whuber Sorry, I'm a little confused as to how you get that. Don't we have $\eta_i = e^{-\lambda t_i}$? Would this get me a different answer to the inconsistent one I included in my edited post? $\endgroup$ – user112495 Apr 16 at 14:47
  • 2
    $\begingroup$ When it has no solution, that proves your initial conditions are impossible. A better test of any solution method is to begin with an actual distribution -- that is, a valid pair $(\lambda, T)$ -- and compute two pairs $(x,F(x)).$ Then work backwards to check that you recover $(\lambda,T)$ from that information. $\endgroup$ – whuber Apr 16 at 15:43

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