$X_1,X_2,...,X_n$ and $Y_1,Y_2,...,Y_n$ are independent samples from the exponential distributions with parameters $\lambda$ and $\frac{1}{\lambda}$.

What is the MLE for $\lambda$?

I used the log-likelihood function:

$logL(\lambda)=\mathbb{}log\prod\lambda e^{-\lambda x_i}\prod\ \frac{1}{\lambda} e^{-\frac{1}{\lambda} y_j}=$

$=n\cdot log(\lambda)+n\cdot log(\frac{1}{\lambda})-\lambda\sum x_i-\frac{1}{\lambda}\sum y_j \rightarrow max\lambda$

After differentiation I got:

$\frac{n}{\lambda}+n\lambda-\sum x_i-\sum y_j =0$

And this is a quadratic equation for $\lambda$.

Can someone check my calculations? Is it correct that there are two possible ML estimations for $\lambda$?

  • $\begingroup$ The derivative of $n\log(1/\lambda)$ w.r.t $\lambda$ is $-n/\lambda$, not $n\lambda$. Even simpler is to recognize that the normalization constants $\lambda^n$ and $(1/\lambda)^n$ cancel in the likelihood. Note, too, that the derivative of $-1/\lambda$ is $1/\lambda^2$. $\endgroup$
    – whuber
    Dec 8, 2014 at 23:33

1 Answer 1


Let's assume that $x_i, y_i > 0,$

The formula simplifies to

$$ n\cdot \log(\lambda)+n\cdot \log(\frac{1}{\lambda})-\lambda\sum x_i-\frac{1}{\lambda}\sum y_j \\=n\cdot \log(\lambda)+n\cdot \log(1)-n\cdot \log(\lambda) -\lambda\sum x_i-\frac{1}{\lambda}\sum y_j \\=-\lambda\sum x_i-\frac{1}{\lambda}\sum y_j $$

and the derivative w.r.t. $\lambda$ is $$ -\sum x_i+\lambda^{-2}\sum y_j $$

so if we define $\bar{x} = \sum x_i$ and $\bar{y}=\sum y_j $, then

$$ -\bar{x}+\lambda^{-2}\bar{y} = 0 $$

$\lambda = \sqrt{\frac{\bar{y}}{\bar{x}}}$, since $\lambda$ needs to be positive


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