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What is the standard error of the mean of an exponential distribution of the form $Ae^{Bx}$ with $N$ measurements?

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I will recast your problem in slightly more formal language. Let $X$ have density function $Ae^{Bx}$ for $x \ge 0$, and $0$ for $x<0$. Then $B$ has to be negative, say $B=-\lambda$, where $\lambda$ is positive. And for the integral of the density function from $0$ to infinity to be $1$, you need $A=\lambda$.

What is the mean of $X$? It is a standard fact that this is $$\int_0^\infty x\lambda e^{-\lambda x}dx$$ Integrate (by parts), or if you are in a pre-integration phase, accept the fact that this integral is equal to $1/\lambda$.

What is the variance of $X$? By a standard fact, this is $E(X^2)-(E(X))^2$. For $E(X^2)$, integrate $x^2$ times your density function from $0$ to infinity. After a while, you will find that the variance of $X$ is $1/\lambda^2$. Or else accept from your notes that this is the case.

Now let $X_1, X_2,\dots,X_N$ be independent exponentially distributed all with the same $\lambda$. Your sample mean is the random variable $$Y=(X_1+X_2+\cdots +X_N)/N$$ Thus $Y$ is a constant ($1/N$) times a certain sum. The variance of $Y$ is $(1/N^2)$ times the variance of $X_1+X_2+\cdots +X_N$. But the variance of a sum of independent random variables is the sum of the variances, so the variance of $Y$ is $$(1/N^2)(N/\lambda^2)$$ which is $1/(N\lambda^2)$. For the standard error, aka standard deviation, of $Y$, take the square root of the variance. You will get $$\frac{1}{\lambda\sqrt{N}}$$ You may have to adapt the reasoning to the particular tools, and language, that you are expected to use. I hope that will not be difficult.

Note: Maybe your notes/text just tell you that the standard error of the exponential is $1/\lambda$, and that the standard error of a sample mean is $1/\sqrt{N}$ times the standard error of any one of your experiments. Then the answer $1/(\lambda\sqrt{N})$ comes with essentially no calculation needed.

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Thanks for your informative answer. I have fit some data to the exponential form I gave before, and B and A are not the same. Is it still possible to use your solution? It appears to me that I do not have a single value $\lambda$ to work with. – shino Apr 30 '11 at 18:01
@shino: There are generalizations of the exponential distribution, sometimes also, confusingly, called the exponential, in which one starts at some place $a \ne 0$. Also, some generalizations have $A$, $B$ varying instead of constant! But your problem may be either (i) no exponential fits your data very well or (ii) the fitting is being done incorrectly. Is there some good physical reason not to start at $0$? – André Nicolas Apr 30 '11 at 18:13
@shino: $A$ and $B$ are not supposed to be the same, $B$ is negative, and $A=|B|$. – André Nicolas Apr 30 '11 at 18:19
@shino: Another thought. Are you trying to fit an exponential to sample means? Sample means from an exponential distribution do not have exponential distribution. – André Nicolas Apr 30 '11 at 18:58
@shino: Or else if you are doing everything correctly, and exponential is a poor fit, look for a better fit from one of the Weibull distributions. It is not hard to find information on how to do this in standard stats books, also presumably the Internet. – André Nicolas Apr 30 '11 at 19:09

$Ae^{Bx}$ isn't a probability density (supported on the positive real numbers) unless it is non-negative and integrates to $1$ which only happens when $B < 0$ and $A = |B|$, which is the "classical" exponential distribution and the answer of user6312 is apt.

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