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1h
comment Bounding a strange function
@amathnerd Why upvote a post not answering the question?
1h
comment To evaluate integral using Beta function - Which substitution should i use?
$$t=\frac{(a+b)x}{a+bx}$$
1h
comment Expectation of scaled sum of squares of iid random variables
For every $i\ne j$, the distributions of $(X_i,\|X\|)$ and $(X_j,\|X\|)$ coincide hence $E(X_i^2\|X\|^{-2})=E(X_j^2\|X\|^{-2})$. Summing these from $i=1$ to $n$ yields the expectation of $(X_1^2+\cdots+X_n^2)\|X\|^{-2}=1$ hence $E(X_i^2\|X\|^{-2})=n^{-1}$ for every $i$ and you are done.
1h
comment If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?
Nobody "moves things under integrals",as you keep insisting on doing. Everything I wrote is justified. Enough is enough.
4h
comment If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?
Nothing breaks down, thank you. Once again (but for the last time), of course $X_tdt+dB_t=dV_t$ for $V_t=\int_0^tX_sds+\int_0^tdB_s$ hence $A_tdV_t=A_tX_tdt+A_tdB_t$, that is, $\int_0^tA_sdV_s=\int_0^tA_sX_sds+\int_0^tA_sdB_s$.
4h
comment Expectation of scaled sum of squares of iid random variables
By symmetry, the result is exactly $$\color{red}{\bf\frac{k}n}.$$
9h
comment If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?
Of course $X_tdt$ is some $dV_t$, for $V_t=\int_0^tX_sds$. Likewise, $X_tdt+dB_t$ is $dV_t$ for some process $(V_t)$, for $V_t=\int_0^tX_sds+\int_0^tdB_s$. These are the definition of the differential notations in SDE.
17h
comment If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?
See previous comment (did you actually read it?).
18h
comment If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?
It is quite valid to multiply the SDE in its differential form by any factor one wants. Recall that $dU_t=dV_t$ means that $U_t-U_0=V_t-V_0$, then $A_tdU_t=A_tdV_t$, that is, $\int_0^tA_sdU_s=\int_0^tA_sdV_s$. On the other hand, what you did is equivalent to state that $A_tX_t=A_0X_0+\int_0^tA_sdX_s$, which might not hold, already for integrals with no stochastic part.
19h
comment If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?
The author probably rather suggests to compute $$d(e^{-t}X_{t})= e^{-t}dX_{t}-e^{-t}X_tdt=e^{-t}(X_tdt+dB_t)-e^{-t}X_tdt=e^{-t}dB_t$$ hence $$e^{-t}X_t=X_0+\int_0^te^{-s}dB_s.$$
21h
comment Dermining stable and unstable manifolds - is my result ok?
Yeah, once again this is a 1D problem hence all this talk about stable and unstable manifolds is rather offtopic in the present case, don't you think?
21h
comment Given probability distribution $f(x)=2-bx$ find $b$ and range for $x$
Yes, $0<c\leqslant1$ and $b(c)=2(2c-1)/c^2$ hence $b(c)\leqslant b(1)=2$. To get $b(c)>0$, use $1/2<c\leqslant1$.
1d
comment Dermining stable and unstable manifolds - is my result ok?
Stable and unstable manifolds of equilibria of 1D differential equations? What for?
1d
comment Given probability distribution $f(x)=2-bx$ find $b$ and range for $x$
Actually, any interval $(0,c)$ with $c\geqslant1$ supports the PDF $f_c:x\mapsto f_c(x)=(2-bx)\mathbf 1_{0<x<c}$, for $b=2(2c-1)/c^2$. Thus, the solution in this post, which is simply $f_1$, is far from being unique.
1d
comment The “how many pieces do you have buy on average” problem, a markov problem?
Yes the number of different sets one gets after one bought k sets is a Markov chain on {0,1,...,n} with transitions from k to k and from k to k+1 only.
1d
comment Given probability distribution $f(x)=2-bx$ find $b$ and range for $x$
You are given that $f>0$ only on some range. Why do you see fit to integrate $f$ on $(0,\infty)$?
1d
comment Transitivity of a binary relation on the power set
The relation E={{1,2},{3,4}} already disproves the first point, it seems, since {1} and {2,3} are related by F and {2,3} and {4} are related by F but {1} and {4} are not.
1d
comment Why is $E(X_2|X_1) = X_1$?
You are actually computing $E(X_2\mid X_1=x_1)$ as being $x_1$ for every $x_1$ in $(0,1)$, hence indeed $E(X_2\mid X_1)=X_1$ almost surely.
1d
comment If $P(X_1 < X_2)$, what is $P(X_1 < X_2 \cap X_1 < X_3)$?
Hint: Can you compare $P(X_1 < X_2 \cap X_1 < X_3)$, $P(X_2 < X_1 \cap X_2 < X_3)$ and $P(X_3 < X_1 \cap X_3 < X_2)$? What is their sum? (Note that this is reminiscent of the case of two random variables, which can be solved by comparing $P(X_1 < X_2)$ and $P(X_2 < X_1)$ and computing their sum.)
1d
comment Poisson Process with continuous rate, Finding Conditional Number of Arrivals
Why repost the exact same question? To game the system? The trouble is that, now, posting the solution I reached would be to condone this behaviour...