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Hot answers tagged probability

3

You missed $\pmatrix{0&1\\1&0}$ in the $n=2$ case, so the correct count in that case is $10$. To find the count for $n=3$, first count the matrices without identical rows: There are $8\cdot7\cdot6$. Then count the matrices without identical rows but with identical columns. Say the first two columns are identical. Then there are four different ...

2

As it was hinted in the comments, the result is not true unless you assume uniform convergence somewhere. Here is a simple example. Let $X_n = X$ have non-degenerate Bernoulli distribution and $Y_{un} = X \mathbf{1}_{u\ge n}$. Then $P(Y_{un}\neq X_n)\to 0$, $u\to\infty$, and $e^{itY_{un}} \to 1$, $n\to\infty$, in all senses. However, $e^{itX_n} = e^{itX}\... 2 Yes, it's a reasonable definition. The problem you seem to be having, however, is why it works for continuous and not for discrete. We can look at this for two cases: If the random variable is discrete and$c = d$, then$P(c \leq X \leq d)$for$c = d$would just be$P(X=c)$, which could be some non-zero value. The integral, however, would be$\int_{c}^{...

2

To make things slightly simpler, say there are $n$ balls in the box. Tge expected value of the first ball is (n + 1)/2. Suppose we've drawn k balls already and the maximum so far is M. The expected value of the next maximum is $$E(M_{next}) = M^2/n + (M+1)/n + (M+2)/n + \ldots + n/n$$ This gives you a recurrence relation for $E(M_k)$ that you can ...

2

The right answer is $89.04\%$. One may write \begin{align} P(0.610<X\leq0.618)&=P\left(\frac{0.610-0.6140}{0.0025}<\frac{X-0.6140}{0.0025}\leq\frac{0.618-0.6140}{0.0025}\right) \\&=P\left(-1.6<Z\leq 1.6\right) \\&=\Pi\left(1.6\right)-\Pi\left(-1.6\right) \\&=2\times\Pi\left(1.6\right)-1 \\&=0.8904\cdots \end{align} giving $... 1 We give a very weak result: sometimes the only functions$g$are the trivial ones. Let$X$take on values$0$and$1$, and let$\Pr(X=1)=p$, where$p$is transcendental. Let$g(0,0)=a$,$g(0,1)=b$,$g(1,0)=c$,$g(1,1)=d$. Here$a,b,c,d$only take on values$0$and$1$. The corresponding probabilities are$(1-p)^2, p(1-p), p(1-p),p^2$. There are two ways the ... 1 No, there is no such random variable$\xi$and constant$c \neq 0$so that$\xi$and$\xi + c$are distributed the same. Why? If these two random variables were distributed the same, then their CDF's (say$F_1$and$F_2) would also be the same. Now $$F_1(z) = P(\xi \leq z) = P(\xi + c \leq z + c) = F_2(z+c) = F_1(z+c).$$ Therefore, the CDF is periodic ... 1 Yes, in a finite case like this the expected value and variance exist; you don't need to prove that in detail. You can calculate the expected value from the expression you wrote, but quite often, including in this case, it's easier to calculate it like this: \begin{align} E(M)&=\sum_{m=0}^{48}P(M\gt m) \\ &= \sum_{m=0}^{48}\left(1-P(M\le m)\right) \... 1 Why does0!$exist? The ultimate reason that$0!$is allowed to exist is because mathematicians define it to exist. We simply state$0!=1\$ and continue from there. There are reasons why we do this, which others have expounded upon elsewhere, but ultimately there is nothing stopping us from defining whatever we like. You only see the useful definitions ...

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