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Fix some measurable space, $(\Omega,\mathcal{F})$ Suppose that we have a continuum of i.i.d. random variables $\{X_i\}_{i\in[0,1]}$ distributed according to cdf, $G(\cdot)$ (denote the corresponding pdf by $g(\cdot)$). Define $X:=\int_0^1X_i(\cdot)d\lambda(i)$, where $\lambda(\cdot)$ is the Lebesgue measure on $\big([0,1],\mathcal{B}([0,1])\big)$.

Question: What is the distribution of X?

Notes:

  • For simplicity, I assume that the $G$ (and $g$) are either only continuous or only discrete probability distributions (eg, all of the $X_i$'s are iid $\mathcal{U}([0,1])$ rv's). If there is a particular $G$ that is easy for you to explain, please feel free to suggest one.

Thanks in advance!

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    $\begingroup$ I'm not sure such an integral is well defined. If $X_i$ are Bernoulli 0/1 then why should we expect the set of all $i \in [0,1]$ for which $X_i=1$ to be measurable? The only case I can see for which the integral makes sense is if all random variables are (surely) the same constant $c$. $\endgroup$
    – Michael
    Jun 11 '19 at 3:48
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If $(\omega,i)\mapsto X_i(\omega)$ were jointly measurable (and bounded, say), then you could (i) do the integration, and (ii) $X$ so defined would be a random variable. But this can't be unless the $X_i$ are degenerate. To see this, assume the joint measurability (on top of the iid) and compute the variance of $X$. You'll get $0$, which will force you to conclude that the $X_i$ were degenerate all along.

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  • $\begingroup$ Thanks John and Michael for your answers and apologies for the lag. $\endgroup$ Jul 1 '19 at 21:33

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