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If X and Y are independent, then are X and Y^X independent?

Does the realisation of X have to be the same as the X in the power of Y?

I think this question sounds silly but I'm trying to clear a major misconception in my head. I've always thought that the standalone X and the X in the power of Y could take values from the X distribution independent of each other. I'm asking if standalone X takes a particular value, does power X also have to take it?

FYI, the context: Trying to solve a question about E(KX^(K-1)), where K and X are independent.

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$X$ and $Y^X$ are not independent variables in general. $Y^X$ is a function of the variable $X$ and thus depends on the distribution of $X$. Another way to think about independence is about information, knowing the value of $Y^X$ would give you information about the value of $X$.

Re: misconception. The way you have written it, $X$ in $Y^X$ is the same random variable and take the same value for any realization of the sample. Remember that $X:\Omega \to \mathcal{R}$ is a function from the sample space to real numbers. If we denote the realized value of $X$ as $X(\omega) = x$ for some outcome $\omega \in \Omega$ then for a particular outcome we would have the pair $(X(\omega), Y(\omega)^{X(\omega)}) = (x, Y(\omega)^x)$.

Sometimes it is convenient to define an "independent copy" of a random variable, which is the concept you describe, a independent random variable with the exact same distribution. Usually this is denoted $X^\prime$ and defined explicitly. In the context of your problem, you do not have an independent copy the $X$'s are the same random variable, the same mapping from the sample space to the reals.

To solve your problem, one approach would be the law of iterated expecatations

\begin{align} E[KX^{K-1}] &= E[E[KX^{K-1}|K =k]]\\ &= E[KE[X^k|K=k]] \end{align}

Once we condition on the value of $K$, the inner expectation is just a function of $X$ and a constant. Since the variables are independent, conditioning on the even $K=k$ does not give us information about the distribution of $X$. Thus the inner expectation can be solved using only the pdf of $X$ and the law of the unconscious statistician. The other way to solve this is to just directly calculate the expecation over the joint distribtion of $X,K$. The joint distribution is just the product of the marginal pdfs due to independence. Solving the integral in practice will mirror the law of iterative integration approach (integrate over marginal of $X$ holding $K$ constant, then integrate the result over the marginal of $K$, or of course vice-versa. Notice this is exactly what the iterative approach does.)

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  • $\begingroup$ this is amazing! thank you so much! :) $\endgroup$ – Ian Petrus Tan Oct 18 at 8:30
  • $\begingroup$ Thanks. if the answer is satisfactory, please mark it as the chosen answer. $\endgroup$ – Tyrel Stokes Oct 18 at 18:24

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