High School Is the Mean Using the CDF Calculation Correct?

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The discussion focuses on calculating the mean of the function e^{-X} using the random variable X's cumulative distribution function (CDF) and probability density function (PDF). The mean is expressed as an integral involving the PDF, but due to the complexity of X, the PDF is difficult to derive. An alternative approach using integration by parts is suggested, leading to a reformulation of the mean in terms of the CDF. The main inquiry is whether this reformulation is mathematically valid. The consensus is that the approach appears correct.
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Hello all,

I have the random variable ##X## with CDF and PDF of ##F_X(x)## and ##f_X(x)##, respectively. Now I have a function in terms of the random variable ##X##, which is ##e^{-X}##, and I want to find the mean of this function. Basically this can be found as

\int_0^{\infty}e^{-x}f_X(x)\,dx

where the range of the random variable ##X## is between ##0## and ##\infty##. However, ##X## is a very complicated random variable (it's a function of a number of other independent random variables), and thus although the CDF is easy to find relatively, the PDF is not that nice expression after taking the derivative of the CDF function. So someone here pointed out to me that by using integration by parts, the above mean expression can by written as

\int_0^{\infty}e^{-x}f_X(x)\,dx=\left. -e^{-x}F_X(x)\right|_0^{\infty}+\int_0^{\infty}e^{-x}F_X(x)\,dx=\int_0^{\infty}e^{-x}F_X(x)\,dx

I just wanted to make sure: Is this mathematically correct?

Thanks
 
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Anyone on this?
 
Looks good to me.
 
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