I am having a random dataset which seems to have exponential distribution. It is generated using the formula given in the answer here
x = log(1-u)/(−λ)
I am trying to reverse engineer, and trying to find out the rate parameter used in generating the data set. Can I calculate λ based on the set of data available?
Yes. Given a random sample $\boldsymbol x = (x_1, \ldots, x_n)$ drawn from an exponential distribution with rate parameter $\lambda$, namely $$f_X(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ we observe that the likelihood given the sample is $$\mathcal L(\lambda \mid \boldsymbol x) = \prod_{i=1}^n f_X(x_i) = \lambda^n e^{-\lambda n \bar x} \mathcal{1}(x_{(1)} > 0),$$ hence the log-likelihood is $$\ell(\lambda \mid \boldsymbol x) = n \log \lambda - n \bar x \lambda.$$ The MLE is therefore $$\hat \lambda = 1/\bar x.$$ So all you would do is take the reciprocal of the arithmetic mean of your sample, and this would be your estimate of the rate parameter.