# Tag Info

## New answers tagged uniform

0

The Jeffreys prior for $\theta$ doesn't depend upon the indicator function, although of course the posterior will. The square root of the second derivative of the log likelihood function is all you need: $p(\theta) = \left(-\frac{\text{d}^2(\log \theta)}{\text{d}\theta^2}\right)^{1/2}$ When moving on to the posterior, you'll have to remember that ...

4

In general, the conditional pdf of $X$ given that $X \leq a$ is just $$f_{X \mid \{X \leq a\}}(x) = \begin{cases} \displaystyle \frac{f_{X}(x)}{P\{X \leq a\}}, & x \leq a,\\0, &x > a,\end{cases}$$ that is, it is just the pdf of $X$ scaled to have total area $1$ (as all pdfs must have) in the region of the conditioning event, and $0$ in the ...

3

If the problem does not state explicitly that $X$ and $Y$ are independent, then it doesn't have a solution, because the marginal distributions of $X$ and $Y$ do not determine their joint distribution. Supposing that $X$ and $Y$ are independent, then $e^X$ and $Y$ are also independent. Proof:  \begin{eqnarray} P\left\{e^X \in A, Y\in B \right\} ...

5

Not to suggest that there's anything lacking from Sven's excellent answer, but I wanted to present a relatively elementary take on the question. Consider plotting the two components of each product in order to see that the joint distribution is very different. Note that the product tends only to be large (near 1) when both components are large, which ...

20

It may be helpful to think of rectangles. Imagine you have the chance to get land for free. The size of the land will be determined by (a) one realization of the random variable or (b) two realizations of the same random variable. In the first case (a), the area will be a square with the side length being equal to the sampled value. In the second case (b), ...

2

The number of connected components won't just depend on n, but d as well (if that's something can be varied). Anyway, your best bet is probably to generate a large number of random graphs using the same parameter n (and d) and use this to estimate the probability of getting a single connected component for that n. Vary the magnitude of n and repeat the ...

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