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I'm trying to find the function which allows me to find lambda if:

$$y = 1-e^{-\lambda x}\,.$$

I tried doing this by:

\begin{eqnarray} y-1 &=& -e^{-\lambda x}\\ e^{-\lambda x} &=& -(y-1)\\ -\lambda x &=& \ln(-(y-1))\\ -\lambda &=& (\ln(-(y-1))) / x\\ \lambda &=& -(\ln(-(y-1))) / x \end{eqnarray}

But to my knowledge this doesn't work if $y-1$ is positive.
Have I done this algebra correctly?

Glen_b
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Alesi Rowland
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1 Answers1

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Since you appear to be working with a cdf, keep in mind that $y$ -- being a probability -- must lie between 0 and 1, and being a continuous variate, it has 0 probability of taking any specific value. (You don't need it to be a cdf, specifically, you can verify that $y$ never exceeds 1 in any case, but it's important to keep properties of cdfs in mind if you're using them.)

Your basic approach is okay, though it could be written in a simpler way than you have it.

Glen_b
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  • If you had the time I'd appreciate an example of simpler writing? (bad with algebra but desperately trying to learn). – Alesi Rowland Oct 14 '19 at 08:19
  • An example would be $1-y$ instead of $-(y-1)$; it's more concise. It's fine to leave it as $-(y-1)$ for a few steps if you think it's going to simplify (e.g. by the availability of another minus sign in a few steps) or if you particularly need to keep a (y-1) term visible. – Glen_b Oct 14 '19 at 08:31
  • I'd also encourage you to define your symbols and put in the domain for the various quantities. It would be worth reading the MathJax tutorial & quick reference so you can make your mathematics look how you want it. – Glen_b Oct 14 '19 at 08:39
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    I have made a few edits to your question. Please check it looks like the way you want it to. – Glen_b Oct 14 '19 at 08:46
  • Thanks everyone for all the suggestions! Highly helpful. – Alesi Rowland Oct 14 '19 at 10:22