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Calculus/Density functions
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Question
The standard deviation for a random variable with probability density function f and mean mu is defined as:
sigma={integral(from negative infinity to infinity):[(x-mu)^2]f(x)dx}^(1/2)
Find the standard deviation for an exponential density function with mean mu.
Helpful information:
Normal Distribution
The probability density function of the random variable X is a member of the family of functions:
f(x)=[1/sigma(2pi)^(1/2)]e^{[-(x-mu)^2]/(2sigma^2)}
where sigma is the standard deviation and mu is the mean for the function.
From the definition of the exponential, mu and sigma are 1/lambda.
To compute it, use your definition with the exponential distribution. It requires u-v calculus where ∫v du - uv - ∫v du.
This needs to be done twice, if I remember right, where u is set to the powers of x. You might have to actually work out (x-mu)² to be
x² - 2x*mu + mu² and then divide it into three integrals.
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