Solved – Likelihood ratio test for comparing two exponential distributions

I am trying to use a likelihood ratio test to compare the parameters of two exponential distributions.

by this thread

Likelihood Ratio for two-sample Exponential distribution

I found that I can use

frac {L_{H_1}(hat theta_1,,hat theta_2)}{L_{H_0}(hat theta_0)} = frac {(hat theta_0)^{n_1+n_2}}{(hat theta_1)^{n_1}(hat theta_2)^{n_2}}=left(frac {hat theta_0}{hat theta_1}right)^{n_1} cdot left(frac {hat theta_0}{hat theta_2}right)^{n_2}

to calculate the likelihood ratio statistic.

(with mn1/mn2 being the means of the two datasets)

However, something doesn't seem right to me…I keep getting the statistic = 1 when i plug it into r.

Is this accurite? or does anyone know a likelihood ratio function to compare two distributions in r?

The answer from the link provides a formula of ratio of likelihoods of the null and the alternative hypotheses with detailed derivation, and the R code below is my implementation for it with $theta_1 = 1$, $theta_2 = 2$, $n_1 = 70$, and $n_2 = 100$.

To generate samples, get cdf for exponential distribution first. If $f(x) = frac{1}{theta}e^{-frac{x}{theta}}$, F(x) = $int_{0}^{x} frac{1}{theta}e^{-frac{t}{theta}} dt = 1 – e^{-frac{x}{theta}}$. If $u = 1 – e^{-frac{x}{theta}},$ then $x = -theta*ln(1-u)$. So get $n_1$ samples $u in [0, 1)$ and compute $x_i$ values with $theta = theta_1$ and $n_2$ samples $in [0, 1)$ and compute $y_i$ values with $theta = theta_2$, and then follow the likelihood ratio formula to compute the ratio. The computed ratio is not always 1 using the formula after testing a few times.

theta_ln_1Minusu <- function(theta, vec){ size = length(vec); for(i in 1:size){ vec[i] = -theta*log(1-vec[i]); } return (vec); }

>theta1 = 1; theta2 = 2; n1 = 70; n2 = 100; >u1vals = runif(n1); u2vals = runif(n2);  > xvals = theta_ln_1Minusu(theta1, u1vals); > yvals = theta_ln_1Minusu(theta2, u2vals); > x_avg = sum(xvals) / n1; y_avg = sum(yvals) / n2;  > x_avg [1] 1.041831 > y_avg [1] 1.733426 > w1 = n1/(n1+n2);w2 = n2/(n1+n2); > likelihoodRatio = (w1+w2*y_avg/x_avg)^n1*(w1*x_avg/y_avg+w2)^n2; > likelihoodRatio [1] 168.8613 

On the comment below:

In addition, regarding EngrStudent's comment, I spent a while to compute and found that the ratio is close to 1 only when x_avg is close to y_avg; in other words, $theta_1$ is close to $theta_2$. Denote $frac{yavg}{xavg}$ as $x$, $x > 0$, divide the last term by $(n_1+n_2)$, and then compute the log of the last term in the link:

$w_1*ln(w_1+w_2x) + w_2*ln(w_1/x+w_2) = (w_1+w_2)*ln(w_1+w_2x) – w_2*ln(x) =ln(w_1+w_2x) -ln(x^{w_2})$

If this log value is 0, then $w_1+w_2x=x^{w_2}$. It is known that this equality holds when x = 1. The both sides are continuous. Now check the derivatives of both sides. The left side's is $w_2$, and the right side's is $w_2*x^{w_2-1}$. The increment rate of the left side is fixed to $w_2$, and $ w_2< 1$. If $x < 1$, $x^{w_2-1} > 1$, the increment rate of the right side is always larger than that of the left, $w_2$, and the left is larger than the right when x>0 close to 0. And if $x > 1$, the increment rate of the right side is always smaller than $w_2$. Thus, the equality holds only when $x = 1$.

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