# 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.

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?

Contents

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  1.041831 > y_avg  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  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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