If I have a logit transformed score (was % before transformation), can I transform it back to % without any other info?

[julie2000]

Maybe a really basic question...

If I have a logit transformed score (was % before transformation), can I transform it back to % without any other info?
The same thing for a natural-log-transformed score (was in count form before), can I transform it back to its count form with no other info?
Thanks in advance!!

 

[BigBeachBanana]

Maybe a really basic question...

If I have a logit transformed score (was % before transformation), can I transform it back to % without any other info?
The same thing for a natural-log-transformed score (was in count form before), can I transform it back to its count form with no other info?
Thanks in advance!!

\(\displaystyle \text{logit}(p) = \log\!\left(\dfrac{p}{1-p}\right)\\\)
Can you solve for [imath]p[/imath] given the left-hand side?

 

[julie2000]

I have a log-transformed estimate (e.g. 0.80) and the corresponding % value for this transformed estimate is apparently 69%. But for the rest of the estimates I only have the log-transformed estimate without the corresponding % (which I'm looking for). I've tried this equation to see if 69% would translate back to 0.80 but since the between parenthesis is negative it doesnt work.

 

[BigBeachBanana]

I have a log-transformed estimate (e.g. 0.80) and the corresponding % value for this transformed estimate is apparently 69%. But for the rest of the estimates I only have the log-transformed estimate without the corresponding % (which I'm looking for). I've tried this equation to see if 69% would translate back to 0.80 but since the between parenthesis is negative it doesnt work.

I got 69%. Can you show your work solving for p?

[math]0.8= \log\!\left(\dfrac{p}{1-p}\right)\\[/math]

 

[julie2000]

To see if 69% would transform back I tried: log(69/1-69) which says "error" but I might be doing this wrong.
I'm not sure how to go from 0.8 to 69 I think I'm missing a few steps.
For instance, if the log-transformed estimate is 1,87 how do I get to the original scaled %?
Thanks!!

 

[limiTS]

69% is really 0.69. Try calculating [imath]\displaystyle\ln\left(\frac{0.69}{1-0.69}\right)[/imath]

You should get 0.8

For 1.87, do the following:

[imath]\begin{aligned} 1.87&=\ln\left(\frac{p}{1-p}\right)\\ e^{1.87}&=\frac{p}{1-p}\\ e^{1.87}(1-p)&=p\\ e^{1.87}-e^{1.87}p&=p\\ e^{1.87}&=p+e^{1.87}p\\ e^{1.87}&=p\left(1+e^{1.87}\right)\\ \frac{e^{1.87}}{1+e^{1.87}}&=p\\ \end{aligned}[/imath]

 

[BigBeachBanana]

Or in general...
[math]p=\frac{e^{\text{logit(p)}}}{1+e^{\text{logit(p)}}}[/math]

 

[julie2000]

69% is really 0.69. Try calculating [imath]\displaystyle\ln\left(\frac{0.69}{1-0.69}\right)[/imath]

You should get 0.8

For 1.87, do the following:

[imath]\begin{aligned} 1.87&=\ln\left(\frac{p}{1-p}\right)\\ e^{1.87}&=\frac{p}{1-p}\\ e^{1.87}(1-p)&=p\\ e^{1.87}-e^{1.87}p&=p\\ e^{1.87}&=p+e^{1.87}p\\ e^{1.87}&=p\left(1+e^{1.87}\right)\\ \frac{e^{1.87}}{1+e^{1.87}}&=p\\ \end{aligned}[/imath]

Thank you very much!!