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My pdf for $X$ is $$f_X ( x ; \theta ) = (1/\theta) x \exp (-x/\theta).$$

I am unsure as to how to show that the mean value of $X$ is $\theta$?

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Actually, since he is supposed to show that the mean is $\theta $, I think the error is the extra x in the function. If the function was $\frac{1}{\theta }{e^{ - \frac{x}{\theta }}}$, you could say that this is exactly the function for an Exponential random variable with $\lambda = \frac{1}{\theta }$. Thus, since the expectancy of an exponential random variable with parameter $\lambda $ is $\frac{1}{\lambda }$, the expectancy (or mean) is $\frac{1}{{\frac{1}{\theta }}} = \theta $

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Actually, there is a small error in your expression for the pdf (since it does not integrate to $1$); it should be the pdf of a Gamma$(2,\theta)$ random variable, meaning that $$ f(x;\theta ) = \frac{1}{{\theta ^2 }}xe^{ - x/\theta }, \;\; x>0. $$ Now, the corresponding mean value is $$ \int_0^\infty {xf(x;\theta )\,dx} = 2 \theta $$ (which can be calculated using integration by parts).

EDIT: Since a Gamma$(2,\theta)$ random variable is distributed as the sum of two (independnet) exponential$(\theta)$ rv's, having pdf $\theta ^{ - 1} e^{ - x/\theta }$, $x>0$, each, the expectation can be found by $\theta + \theta = 2\theta$.

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Well, given a continuous pdf, how do you find the expected value (aka the mean)?

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I am unsure of this. I think I am looking for the first expected value is that right? – user5324 Jan 10 '11 at 15:38

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