# Tagged Questions

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### {Probability}: choosing keys from a pool without replacement

The OP is trying to understand the following question. The OP understand that if you can always write out the term $$P(X=k) \implies (1-\frac{1}{N})(1-\frac{1}{N-1})\cdots(1-\frac{1}{N-k+1}),$$ ...
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### Probability: deviation from the mean

I am having trouble to understand the following. If $S_n=X_1+X_2+......+X_n$, where X_1,X_2 are Bernouli (p). I don't understand this. So you get an intermediate point Constant* sqrt(n). To the ...
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### When is an infinite sequence of integers purely deterministic with no randomness involved?

I see in literature very different descriptions of what is a deterministic system such as: "... a system in which no randomness is involved in the development of future states of the system...>>>" I ...
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### The variance of a simple random walk/process

I've been trying to wrap my head around this for the past day. Please help! Let $\epsilon_i = \pm 1$ with equal probabilities independently for $i=1,...,N$. Then $Z_i = \epsilon_1 + ... + \epsilon_i$ ...
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### Off-lattice Brownian bridges in R^3

Start at a point $(0,0,z_0)$ and take $n$ steps of unit length in a random direction (for each step) in $\mathbb{R}^3$. Let such a walk be valid if the position of the last step, and only the last ...
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### How to check that a sequence of numbers is random? [duplicate]

I have a sequence of numbers like 1,7,22,45,12,96,21,45,65,36,85,14,51,16,18,17,16....65... IS there any formula to check whether the sequence is random or not ? In my case odd numbers are ...
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### Random walk on finite graph

I know that the stationary distribution of a random walk on the graph is given by, (degree of the node)/($2\times$ total number of links in graph). My question is, how do we get this solution?
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### Completeness of random walks in multiple dimensions?

I was reading Artificial Intelligence: Modern Approach (Norvig and Russell), and there was a footnote that really caught my attention. I apologize if the problem is more in the domain of CS than ...