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Let us rewrite the product rule as follows: $$(fg)'=f'g+g'f=\frac{f'}{f}fg+\frac{g'}{g}fg=\left(\frac{f'}{f}+\frac{g'}{g}\right)fg$$ Yours is just the generalization to $n$ factors, but is handled in the exact same way.


The first question is: Under what circumstances is $$ \frac \partial {\partial\theta} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \bullet\bullet\bullet $$ the same as $$ \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac \partial {\partial\theta} \bullet\bullet\bullet \text{ ?} $$ The next question is: Why is $$ \frac \partial {\partial\theta} \...


Yes, your calculation is correct. The second game has much higher variance than the first (although to be pedantic, the units of variance are not dollars, but dollars squared). Intuitively, let's think about the distribution of possible outcomes of each game if both are played at the same expected value. In the first case, the random total payout $X$ can ...


Hint: Consider $(0,1)$ endowed with Lebesgue measure $\lambda$ (restricted to $(0,1)$) and $$X_n(\omega) := n 1_{(0,n^{-1})}(\omega).$$


Recall that if $U\sim\mathsf U(0,1)$ and the distribution function $F_Y$ of $Y$ is absolutely continuous, then the random variable $F_Y^{-1}(U)$ has the same distribution as $Y$. In particular, if $F_Y(y) = (1-e^{-y})\mathsf 1_{(0,\infty)}(y)$ then $$F(Y) = U \iff Y = -\log(1-U). $$ Set $Y_k = -\log(1-X_k)-1$, then $Y_k\stackrel d= Y-1$ where $Y\sim\...

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