Solve for $a$ and $b$ when $\Bbb E(X) = 5/8$. I can pretty easily find that $\Bbb E(X)=a/2+b/4$ so $5/8=a/2+b/4$. But then what? Aren't there multiple values for $a$ and $b$? $a=1, b=1/2$, or $a=1/4 b=2$ etc.). How can I solve for $a$ and $b$ from this?
Expectation of $X$ when $f(x)=a+bx^2$, $x \in (0,1)$
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probability
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1I assume $f(x)$ is a probability density function, which by definition, must integrate to --- when you integrate over the domain of $x$. Tell me if you need another hint. – 2012-11-03
1 Answers
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Use the fact that the integral of f(x) between the bounds is equal to one and then substitute equations accordingly.