Alex R.
Reputation
15,754
Top tag
Next privilege 20,000 Rep.
Access 'trusted user' tools
 2h answered Let $f:\mathbb{R}\to[0,\infty)$ measurable and $f\in L^1$. Show that $\mu(E)<\delta \implies \int_E f < \varepsilon$. 4h comment When do we have $E[X_{n+1}\mid X_n] = E[X_{n+1}\mid\mathscr{F}_n]$? @BCLC: You're reading too much into "random variance." You start with two random variables $X_0,X_1$, which will give you physical values after you draw them. You then draw a normal random variable $X_2\equiv N(0,\sigma^2)$ with mean 0 and variance $\sigma^2=X_0+X_1$. The actual variance of $X_2$ is not random but the parameter $\sigma$ is random. Afterall, when you compute the variance of $X_2$, then $\sigma$ comes with some distribution so you'll get a fixed value for the variance of $X_2$. 1d comment simple prime formula $\sum_{i=0}^{\infty}\left\lfloor \frac{2n}{\pi(i)+n+1} \right\rfloor=P_n$ (It's also useless for all practical purposes) 1d answered simple prime formula $\sum_{i=0}^{\infty}\left\lfloor \frac{2n}{\pi(i)+n+1} \right\rfloor=P_n$ 1d comment Problem on divergence, rotation, flux The question is asking for the value of $c$ which makes the equality hold. So start by evaluating both sides for arbitrary $c$. 1d answered Judicious guess for the solution of differential equation $y''-6y'+9y= t^{3/2} e^{3t}$ 2d comment Fourier Transform of operator Yes, the modification is trivial since the multivariate fourier transform is multiplicitive in $e^{-2\pi \xi_ix_i}dx_i$. 2d comment Fourier Transform of operator Are you familiar with how fourier transforms act on derivatives? en.wikipedia.org/wiki/Fourier_transform#Differentiation May 2 comment Prove $A_{\infty} < \infty$? @BCLC: You're right, I'm one step ahead of myself. Just take the limit as $n\rightarrow\infty$. May 2 revised Prove $A_{\infty} < \infty$? added 33 characters in body May 2 answered Prove $A_{\infty} < \infty$? Apr 29 comment How to plot the distribution of a function in r? statmethods.net/graphs/density.html Apr 29 comment Justify the identity $\frac{\partial}{\partial x^i} (\log|\det A|)=b_{rs} \frac{\partial a_{rs}}{\partial x^i}$ When you're looking at $\frac{\partial}{\partial a_{rs}}(det A)$, notice that if you write out the determinant in terms of a cofactor expansion that include $a_{rs}$ in the row/column you're expanding in, then the cofactors are independent of $a_{rs}$, and the only term that survives is the cofactor of $a_{rs}$. Also when chain ruling from $x_i$ to $a_{rs}$ you should have a sum. Apr 28 comment On the linear combination of $\pm 1$ random variables I think you have some typos in your inversion formula. Have you tried plugging in $a=b=0$ into your inversion formula? Then $P(00$ such that $|f(x)-r|<\epsilon$ for all $x\geq N$. Since $N$ is fixed, you can take the limit, and then let $\epsilon$ go to 0. Apr 27 answered Show that $\sin^2 \theta \cdot \cos^2\theta = (1/8)[1 - \cos(4 \theta)]$. Apr 27 comment Show that $\sin^2 \theta \cdot \cos^2\theta = (1/8)[1 - \cos(4 \theta)]$. Your "double angle" formula is the same as the identity you want to prove. Could you correct this? Apr 25 comment LU factorization Algorithm Are you familiar with how similar results are derived for Gaussian Elimination? LU Decomposition is basically the same thing. Apr 24 answered Laplace transforms of powers of cosine Apr 24 answered $\frac{1}{\pi}\int^{\pi}_{-\pi} e^{2x^2} dx\geq\sum_{k=1}^\infty (\frac{1}{\pi}\int^{\pi}_{-\pi }e^{2x^2} \sin(kx)\,dx)^2$