Assume that cube's side has a random length $X$ (= random variable) with uniform distribution for $x \in (0,a)$. I have to determine the expectation of volume and variance of volume.
It is obvious that density of $X$ is a function
$$f_X(x) = \begin{cases}
\frac{1}{a}, & x \in (0,a), \\[2ex]
0, & x \not \in (0,a).
\end{cases}$$
I think I have to determine $V$, where $V$ is a random variable $V = X^3$, am I right? Then I am able to compute it using $\text{E} V = \int_0^ax\cdot f_V(x) \text{ dx}$. I don't know how to determine $f_v(x)$ function. I thought of something like
$$ F_V(x) = P(V \le x) = P(X^3\le x)=P(X\le x^{1/3})=F_X(x^{1/3}) $$
but it doesn't work since (according to the result in my textbook) the $\text{E}V$ should equal $\frac{a^3}{4}$. So how am I suppose to get the result?
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Speedding
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2 Answers
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Since $g(x)=x^3$ is a measurable function, $EV=Eg(X)=\int_0^a g(x)f_X(x)dx$, which gives you the solution.
Javi
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1Well, that sounds obvious but it didn't come to my mind. Thank you very much for clarification! – Speedding Dec 27 '17 at 11:48
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You're welcome! – Javi Dec 27 '17 at 11:54
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You could also approach it in your way.
$$f_V(v) = f_X(g^{-1}(v))\left||\frac{d}{dv}g^{-1}(v))\right|$$
$$f_V(v) = f_X(v^{\frac{1}{3}}) \left|\frac{d}{dv}v^{\frac{1}{3}}\right|$$
$$ = \frac{1}{a} .\frac{1}{3} v^{-\frac{2}{3}}, 0\le v\le a^3$$
$$E(V) =\int_{0}^{a^3} \frac{1}{a} .\frac{1}{3} v.v^{-\frac{2}{3}} = \frac{a^3}{4}$$
Satish Ramanathan
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