Enforcing conditions on truncated exponential distribution

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, 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 $$$$\frac{F(5)}{F(2)} = \frac{1-e^{-5\lambda}}{1-e^{-2\lambda}} = \frac{19}{16}$$$$.

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

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

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?

• 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 $T.$ – whuber Apr 16 at 13:43
• 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. – kjetil b halvorsen Apr 16 at 14:02
• @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. – user112495 Apr 16 at 14:30
• @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? – user112495 Apr 16 at 14:47
• 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. – whuber Apr 16 at 15:43