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I am reviewing an old exam for practice and I would like know if I am on the right track.

The question is

"For every exponential random variable with mean donated by theta, the mode of the random variable is less then mean of the random variable, which in turn is less than the median of the random variable. True? or False?"

because of the shape of the negative exponential curve, that I have used most of often in probability which converges to 0, I assumed True. Because the hight value of the function is 0 and the mean is being weighted, or skewed, to the right but left of the median. Therefore mode

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No. The mode is $0$, which is less than the mean $\theta$, but the median is $\theta\ln 2$, which is less than the mean.

The median can be found as follows. Let the median be denoted by $m$, then

$$ 1-e^{-m/\theta} = \frac{1}{2} $$ $$ e^{-m/\theta} = \frac{1}{2} $$ $$ e^{m/\theta} = 2 $$ $$ m/\theta = \ln 2 $$ $$ m = \theta \ln 2 $$

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