# Back transforming the intercept of double log normal distribution

I apologise if this is a duplicate, i couldn't find a thread that seem to be talking about the same thing.

I have a dataset with a bunch of duration of varying lengths in seconds, i log-transformed the dataset and binned them in to bins with width of 0.5 log-units. using this data, i plotted the following bar graph.

Next, I obtained the maximum likelihood estimation of the of my data set

p = 0.856

$$\mu_1$$ = 0.831

$$\mu_2$$ = 5.903

$$\sigma_1$$ = 1.036

$$\sigma_2$$ = 0.703

using these parameters, i drew the double gaussian according to

$$y = p(\frac{1}{\sigma_1\sqrt{2\pi}}exp(-\frac{(x-\mu_1)^2}{2\sigma_1^2})) + (1-p)(\frac{1}{\sigma_2\sqrt{2\pi}}exp(-\frac{(x-\mu_2)^2}{2\sigma_2^2}) )$$

2 distributions will cross at $$(x_c,y_c)$$ when

$$p(\frac{1}{\sigma_1\sqrt{2\pi}}exp(-\frac{(x_c-\mu_1)^2}{2\sigma_1^2})) = (1-p)(\frac{1}{\sigma_2\sqrt{2\pi}}exp(-\frac{(x_c-\mu_2)^2}{2\sigma_2^2}))$$

$$x_c$$ = 3.25

I wish to back transform $$x_c$$ into seconds

my understanding is that $$exp(x_c)$$ will not give me an accurate estimation. Because if, for example, i wish to back transform the $$\mu$$, $$exp(\mu)$$ gives me the geometric mean instead of the arithmetic mean. a more precise way is to use the Johnson back-transformation which is $$exp(\mu +0.5\sigma^2)$$. Thus, my question is, how do i back transform $$x_c$$ to get an accurate estimation in seconds?

• Your expression is not a "transform" of $x,$ because for any given value of $y$ there may be two and as many as four corresponding values of $x.$ This expression is that of the density of a mixture of Normals, which is rarely (if ever) used for transformation (and has nothing to do with what is usually understood to be a "lognormal" distribution). One has to suppose you are confused about what is happening, but without having any context, we have little guidance for setting you straight. Could you tell us the statistical problem you are trying to solve? – whuber Apr 30 '19 at 12:22
• Thanks for the reply, i am currently trying to apply the same method as detailed in this paper. in short, after log transformation and binning the data into bins with width of 0.5 log units, the probability density of my data show up as a double normal distribution. i wish to obtain the point where the 2 distributions cross – Villager A May 1 '19 at 3:45
• Thank you for the clarification. If I understand it correctly, it means the references to log transformation, binning, and even back-transformation are just distractors and that your question is how to solve the stated equation for $x$ for the given values of the parameters. After taking logarithms of both sides you will obtain a quadratic equation: applying the quadratic formula does the trick. You can therefore anticipate obtaining two solutions for the crossing (which will occur whenever $\sigma_1\ne\sigma_2.$ Pick the one lying between the means $\mu_i.$ – whuber May 1 '19 at 13:36
• Thanks for the answer, i am aware of how to obtain $x$ which is 3.25 log units. My confusion is that since this data has been log transformed, $x$ = 3.25 log units is not exactly of what i am trying to find. I need to back transform $x$ into the original scale. And that is where is am stuck – Villager A May 2 '19 at 3:25
• The "back-transform" of the logarithm is, by definition, the exponential function. – whuber May 2 '19 at 12:37