SOLUTION: Let X be a continuous random variable with density function
f(x) = { theta*x + 3/2*theta^{3/2}*x^2, for 0 < x < 1/sqrt( theta ); 0, otherwise }
where theta > 0. What is the expec
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Question 1203892: Let X be a continuous random variable with density function
f(x) = { theta*x + 3/2*theta^{3/2}*x^2, for 0 < x < 1/sqrt( theta ); 0, otherwise }
where theta > 0. What is the expected value of X?
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!
To find the expected value, we apply the integral to x*f(x) over the limits from -infinity to infinity.
E(x) = expected value
 = \int_{-\infty}^{\infty}\text{x}*f(\text{x})d\text{x})
Because much of the real number line makes f(x) = 0, we only need to worry about the interval which is the same as where theta > 0
The goal has been reduced to computing
 = \int_{0}^{\sqrt{1/\theta}}\text{x}*f(\text{x})d\text{x})
Using fairly elementary calculus integration rules, you should find that:
 = \int \text{x}*f(\text{x})d\text{x})
 = \int \text{x}*(\theta*\text{x} + (3/2)\theta^{3/2}\text{x}^2)d\text{x})
 = \int (\theta*\text{x}^2 + (3/2)\theta^{3/2}x^3)d\text{x})
 = (1/3)\theta*\text{x}^3 + (3/8)\theta^{3/2}\text{x}^4+C)
For some constant C.
Then,
 = (1/3)\theta*0^3 + (3/8)\theta^{3/2}*0^4+C)
 = C)
and
 = (1/3)\theta*(1/\sqrt{\theta})^3 + (3/8)\theta^{3/2}*(1/\sqrt{\theta})^4+C)
 = (1/3)\theta*\theta^{-3/2} + (3/8)\theta^{3/2}*\theta^{-2}+C)
 = (1/3)\theta^{-1/2} + (3/8)\theta^{-1/2}+C)
 = (1/3+3/8)\theta^{-1/2}+C)
 = (17/24)\theta^{-1/2}+C)
 = C + \frac{17}{24\sqrt{\theta}})
Subtract the two results to find
 = g(1/\sqrt{\theta}) - g(0))
 = C + \frac{17}{24\sqrt{\theta}} - C)
 = \frac{17}{24\sqrt{\theta}})
Therefore, the expected value is 
where theta > 0
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