Calculating the Expected Value of a Probability Density Function (steps)

Question

I have this probability density function and I need to find its expected value: $$f(t)=be^{-bt}$$ Which was also given to be: $$E[X]=∫_{-∞}^∞tf(t)dt$$ $$E[X]=∫_{-∞}^∞tbe^{-bt}dt$$ I also know the answer, which is: $$E[X]=\frac{1}{b}$$ And I understand that integration by parts was used, but I don't know how, so if someone could elaborate on the steps involved that would be amazing.

Answer 1

Before deriving you expectation you have to understand which kind of distribution you are facing

$$f_X(t)=b e^{-bt}$$

is the density of a Negative exponential distribution, with $b>0$ and $X \geq 0$ thus your integral becomes

$$\mathbb{E}[X]=\int_0^{\infty} bt e^{-bt}dt=\frac{1}{b}$$

The integral is immediate if you know the definition of Gamma Function

$$\mathbb{E}[X]=\frac{1}{b}\int_0^{\infty} (bt) e^{-bt}d(bt)=\frac{1}{b}\Gamma(2)=\frac{1}{b}$$