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4

Observe that for each $n$, $\{X<\infty\} = \{\sup_{k\ge n}X_k <\infty\}$.


2

Oops, you made a mistake. $P(X\geq1)=1-P(X=0)$. You have it as $P(X\geq1)=1-P(X=0)-P(X=1)$


2

So we have $P(X>Y) = P(X-Y>0) = P ((X,Y) \in C)$ Where $C = \{(x,y):x-y>0\}$ But we know how to find this, as this is the same as simply evaluating f(x,y) and integrating this over the region C. $P(X>Y) = \iint_C f(x,y) dxdy$ .


2

Your model is different. If he buys a ticket with the winning number from the winning lottery, probability that it will appear next time, is $10^{-5}$. Now you have easy to solve inequality $$ \left(1-10^{-5}\right)^n\leq 0.5. $$ (And the answer to your second question is: yes. If choosing the winning ticket is really random, it is meaningless, which one ...


2

What you want is $|cY|=|c|\cdot |Y| \le 1$; knowing that $|Y|\le 1$ and that the boundary can be reached, it is necessary and sufficient to have $|c|\le 1$. So your second answer, $[-1,1]$, is correct.


1

For the last series, since the second is convergent it suffices to check that the series $\sum_{n\geqslant 1}b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]$ is convergent. To this aim, we start writing that $$b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]\leqslant \int_0^1 t\mathbb P\{|b_nX_1|\geqslant t\}\mathrm dt.$$ ...


1

As you say, the explanation is terse. You correctly understand why you can say $4\theta_1 + 6\theta_2 = 1,$ and the formula $2\theta_1 + 3\theta_2 = \frac12$ appears to be, as you say, just dividing that formula by $2$. But another motivation for writing the formula that way is that the sum of probabilities in a single column of the joint probability ...


1

So if I understand right, if $X-\mathbb EX$ is $\sigma^2$-sub-Gaussian, then $$\forall t\in\mathbb R:\mathbb Ee^{t(X-\mathbb E X)}\le e^{t^2\sigma^2/2}\tag1$$ Taking the logarithm of both sides of (1), we have an expression in terms of the so-called cumulant-generating function of $X$: $$\forall t\in\mathbb R:\log\mathbb Ee^{tX}-\mathbb EtX\le ...


1

(a) Your answer is right until the last step. Because $N(t)$ has a Poisson distribution with parameter $\lambda t$, it should be: $$P(N(t)=1)\cdot P(N(t)=1)\cdot P(N(t)=1) = (\lambda t \;e^{-\lambda t})^3.$$ (b) Your answer is right. (c) \begin{eqnarray*} E(T_2\mid T_1\lt T_2\lt T_3) &=& \int_{t_2}{t_2 P(T_2=t_2\mid T_1\lt T_2\lt T_3)\;dt_2} \\ ...


1

Since the sequence $(X_n)_{n\geqslant 1}$ is independent, an application of the Borel-Cantelli lemma shows that $X_n\to 0$ almost surely is equivalent to the convergence of the series $\sum_{n\geqslant 1}\mathbb P(|X_n|>\varepsilon)$ for each fixed $\varepsilon$. Thus, $X_n\to 0$ almost surely is and only if $\sum_{n=1}^\infty\mathbb ...



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