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17h
comment Question about sigma-algebras
Show that $\mathcal{A}=\{B\subseteq\mathbb{R}\mid f^{-1}(B)\in\sigma(f^{-1}(\mathcal{D}))\}$ is a sigma-algebra that contains $\mathcal{D}$.
19h
comment Question about sigma-algebras
By definition $\sigma(f)=f^{-1}(\mathcal{B}^n)$ which is already a sigma-algebra, so no need to write $\sigma(f^{-1}(\mathcal{B}^n))$. Now use that if $\mathcal{D}$ is a generator of $\mathcal{B}^n$, then $f^{-1}(\mathcal{B}^n)=\sigma(f^{-1}(\mathcal{D}))$.
22h
comment Probability normal distribution P(X>Y)?
Are the two variables assumed independent? Otherwise, you can't say much.
22h
comment Calculating Covariance missing understanding for one step
Yes exactly and as consequence they get $\sum_{i=1}^9 (x_i-x_{\mathrm{avg}})=-0.01$ which is obviously wrong since $\sum_{i=1}^9 (x_i-x_{\mathrm{avg}})=0$ always. But the important message is that they use a different formula than the covariance formula.
1d
answered Calculating Covariance missing understanding for one step
1d
answered Question about sigma-algebras
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
Is $\tau-1$ necessarily non-negative?
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
Yes.${}{}{}{}{}$
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
"especially for t=c" - but you just said c=t-1. Anyhow, for $t\geq 1$ we have $\{\tau+1\leq t\}=\{\tau\leq t-1\}\in\mathcal{F}_{t-1}$ and thus also $\{\tau+1\leq t\}\in\mathcal{F}_t$ (why?).
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
Well, you haven't used that $\tau$ is a stopping time yet. What does that tell you? And what is that mysterious $c$ you introduced?
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
If $T=[0,\infty)$, then distinguish between the two cases: $t<1$ and $t\geq 1$.
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
Then please revise your attempt. And is it now obvious that $\tau-1$ is not a stopping time?
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
No, it is $\{\omega: \tau(\omega)\leq t\}\in\mathcal{F}_t$, as the other condition does not make sense. Usually, a stopping time is defined to be non-negative (at least when studying stochastic processes indexed by $[0,\infty)$). Is that the case here as well?
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
Please provide your definition of being a stopping time. Something tells me you haven't got it entirely right judging from your attempt.
1d
comment $\tau$ is stopping time. Check if $\tau + 1$, $\tau - 1$, $\tau^2$ also are stopping time.
$A\subseteq B$ and $B\in\mathcal{F}_t$ does not imply that $A\in\mathcal{F}_t$. If that were the case, then $\mathcal{F}_t$ would contain every set.
1d
answered Inner dependence of Independent random vectors
1d
comment Inner dependence of Independent random vectors
No.${}{}{}{}{}{}$
2d
comment Prove $\mathsf E(N)=\sum_{i=1}^\infty \mathsf P(N\geqslant i)$
Integrate the pointwise equality $N=\sum\limits_{i=1}^\infty \mathbf{1}_{N\geqslant i}$ with respect to $P$. This of course only holds for discrete random variables with values in $\mathbb{N}$.
Oct
22
comment Using Convolution to find density of sum of non-independent normal densities
Convolution requires that the variables are independent.
Oct
20
revised Equivalent definition of random variables
added 17 characters in body