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4

Strictly speaking, at any given point the pdf doesn't mean anything. You can change the value of the pdf at that particular point to be whatever you want, and the distribution of the random variable is the same. Still, "generically", $f(x)$ is essentially the probability to find the random variable in the interval $(x-\epsilon,x+\epsilon)$, divided by $2 ...


4

There are $10$ choices for the digit that appears twice and then $9$ choices for the digit that appears once. There are $\binom{3}{1}=3$ ways to arrange the digits. This gives $10\cdot9\cdot3=270$ choices out of $1000$. That is, a $0.27$ probability.


2

The velocity of a moving particle is the rate of change of the displacement at that time. The probability density of a continuous real valued random variable is the rate of change of the cumulative probability at that point.


2

As I pointed out in a comment, you can choose a single number with linearly increasing probability by choosing two distinct numbers and using the larger one. (This gives probability $0$ for the least element, so you need to add a dummy least element to get exactly the probabilities you describe.) To use this for efficiently selecting $k$ distinct elements ...


2

I finally ended up using weighted reservoir sampling with uniformly increasing weights. Here is a detailed paper: http://utopia.duth.gr/~pefraimi/research/data/2007EncOfAlg.pdf Implementation in Java: http://utopia.duth.gr/~pefraimi/projects/WRS/


2

Your reasoning in the "My understanding" paragraph is right. Why would it be $(1-\alpha)^{10000}?$ You multiply probabilities for independent events, but the events "ex-convict A is guilty" and "ex-convict B is guilty" are not independent. On the contrary, you have correctly identified them as being mutually exclusive, which is about as far from independent ...


1

The pawns need not be on the diagonals. And you have counted the number of way to place the pawns on the diagonals incorrectly, there are only $2$ ways. For remember that $\binom{64}{8}$ counts the number of ways to choose the set of positions of the pawns. So in counting favourables, we must also count sets of positions only. The top left to bottom right ...


1

The player meets monsters at the rate $\lambda$, and with probability $p$ they are winning monsters, so the player meets winning monsters at the rate $\lambda p$. We can extend the game up to $t=1$, with monsters appearing after the player's death meeting a dead player. Then, since the Poisson monsters appear uniformly in time, the probability distribution ...


1

The events represented by the terms of your summation are not pairwise disjoint. Consider, for example, the outcome in which the first $n$ trials are successes; that situation is counted both in the $k=0$ term and in the $k=1$ term, since the latter includes the sequence of $n$ successes followed by $1$ failure. We can also use the fact that $$\sum_{k\ge ...


1

Step 3 is given by $ \mathrm{Var}[X] = \sum p_i x_i^2 + (\mathrm{E}[X])^2 \cdot \color{red}{\sum p_i} - 2 \cdot \mathrm{E}[X] \cdot \color{blue}{\sum p_i x_i}. $ This is equivalent to $ \mathrm{Var}[X] = \sum p_i x_i^2 + (\mathrm{E}[X])^2 \cdot \color{red}{1} - 2 \cdot \mathrm{E}[X] \cdot \color{blue}{\mathrm{E}[X]}. $ This, in turn, is equivalent to ...


1

The field of probability can be made mathematically rigorous. Introductory textbooks on probability tend to use terminology like 'sample space', 'outcome', 'event', and outcomes are given names like 'Heads' or 'HTT' or 'King of Spades', all in an attempt to keep things informal and intuitive. These texts often refrain from defining such concepts precisely, ...


1

Let the result of a game be denoted by $XdY$, meaning that Player $X$ defeats Player $Y$. Then the space of possible results is $$ \{(AdB, AdC), (BdA, BdC), (AdB, CdA, CdB), (BdA, CdB, CdA), (AdB, CdA, BdC, BdA), \ldots\} $$ Partial solution follows. To determine the probability of winning, observe that $C$ wins if and only if the entire series ends ...


1

For part a), $$p(B|D)+p(B^c|D)=\frac{p(B\cap D)}{p(D)}+\frac{p(B^c\cap D)}{p(D)}=\frac{p(B\cap D)+p(B^c\cap D)}{p(D)}$$ $$=\frac{p((B\cup B^c)\cap D)}{p(D)}=\frac{p(D)}{p(D)}=1$$


1

We need to make some assumption about the process of distributing points. I will assume that points are distributed one at a time, that any person is just as likely to get the point at stage $k$, and that the decisions for the various points are independent. Under these assumptions, if random variable $X$ is the number of points Alicia gets, then $X$ has ...


1

UPDATE 2: I managed to prove rigorously that the conjecture I had formulated in the first update holds: The expressions for $\pi_j$ in both formulae are off by a factor of $t$. I also figured out the most plausible reason for this mistake. Consider the partition of the state space $\{S_0,\ldots, S_{t-1}\}$ mentioned above and let $q_{ij}\equiv p_{ij}^{(t)}$ ...



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