New answers tagged

2

Let $(\Omega, \mathcal F, \mathbb P)$ be probability space, let $X: \Omega \to \mathbb R^n$ be random variable. (We consider $\mathbb R^n$ with borel $\mathcal B(\mathbb R^n)$ sigma field.) Moreover, let $f:\mathbb R^n \to \mathbb R$ be arbitrary function and $\mu_X$ the distribution of $X$. Then we have: $$\mathbb E[f\circ X] = \int_\Omega (f \circ X)(\...


-1

The integrand is $1$ on $\{X\leq x, Y\leq y\}$ (which is measurable since $X,Y$ are random variables) and zero elsewhere, so the result immediately follows by the definition of the integral of a simple function.


3

Exchange of order of integration will be an application of Fubini's theorem applied to $g(x,y,\omega)=\Big(\mathbb{1}(X(\omega)\leq x)-\mathbb{1}(X_2(\omega)\leq x)\big)\big(\mathbb{1}(Y(\omega)\leq y)-\mathbb{1}(Y_2(\omega)\leq y)\Big)$. This is $dP\otimes dx\otimes dy$ integrable; in fact, $$\int_\Omega\int_\mathbb{R}\int_\mathbb{R} |g(x,y,\omega)|dx \,dy\,...


2

i) Independence of $(X,Y)$ and $(X_2,Y_2)$ implies independence of $X$ and $Y_2$. This implies independence of $f(X)$ and $g(Y_2)$ for any measurable functions $f,g:\mathbb R \to \mathbb R$. Hence it implies independence of $I_{X_2 \leq x}$ and $I_{Y\leq y}$. Similarly we get independence of $I_{X \leq x}$ and $I_{Y_2\leq y}$. ii) Since $(X,Y)$ and $(X_2,...


2

It is false. Take, for instance, $N=2$, $A=\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$, $B=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$, and $C=\left[\begin{smallmatrix}0&1\\0&1\end{smallmatrix}\right]$.


2

The identity is always false. If you change the left side to $b-a$ then it is always true.


2

We can use the definition of expectation, complete the square, do a change a variable and express it in terms of CDF of standard normal distribution. \begin{align} E[I(X>0)e^X] &= \int_0^\infty e^x \frac1{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\, dx \\ &=\int_0^\infty \frac1{\sqrt{2\pi}} \exp\left( -\frac{x^2-2x}2\right) \, dx \\ &= \int_0^\infty \...


1

Think that $$ 3y^2 \frac{dy}{dx} = 1\Rightarrow\frac{dy}{dx} = \frac{1}{3y^2} > 0\\ 3x^2 \frac{dx}{dy} = 1\Rightarrow\frac{dx}{dy} = \frac{1}{3x^2} > 0 $$ so $y=x^3+1$ is strictly increasing regarding the vertical axis and $x = y^3-2$ is strictly increasing regarding the horizontal axis. Also regarding $y = x^3+1$ it pass across the point $(0,1)$ and ...


0

Using @Andrew's notation let $f(x)=(x^3+1)^3-x-2.$ If expanded there is a single sign change so exactly one positive zero. Now put $x=-2-t$ and expand that, giving a degree $9$ polynomial with all negative coefficients, so $f$ has no zeroes less than $-2.$ [I can list the coefficients if wanted, first few $-1,-18,-144,-669$ via a symbolic calculator I use.] ...


4

Substituting the second equation for $y$ into the first equation, we get $$(x^3+1)^3 = x+2 \iff (x^3+1)^3 - x - 2 = 0.$$ Now consider the polynomial $f(x) = (x^3+1)^3 - x - 2$, for which any root yields a solution to the original system. Thus we want to analyze $f$ in such a way that we can prove there is only one real root. We can see that for any $x$ with $...


1

So the question is, suppose I just want to show above system has a unique solution, but don't care about solving for it Here we are trying to show that the system has only 1 real root. Using the 2 equations, cube the 2nd equation and subtract to get: $$P_9(x)=\:\left(\left(x^3+1\right)^3-x-2\right)=x^9+3x^6+3x^3-x-1=0$$ Applying the Rule of Signs, we ...


1

Note that: In the expression $n^3 = (n^2 - (n-1)) + (n^2 - (n-3)) + ... + (n^2 + (n-3)) + (n^2 + (n-1))$, $n-1$ cancels from the first term and the last term. $n-3$ cancels from the second term and the second to last term. $n-5$ cancels from the thirst term and the third to last term. And so on. $n-a$ cancels from the $(\frac{a+1}{2})^\text{th}$ term ...


1

After cancelling what cancels, you just have $n^3=n^2\cdot n.$ (hopefully helpful) edit: We can apply arithmetic sequence results. If an arithmetic sequence has first term $a,$ common difference $d,$ last term $l$ and number of terms $n,$ then for $1 \le k \le n$ the $k$th term $t_k$ is $a+(k-1)d.$ [so last term $a+(n-1)d.$] The sum is then $(1/2)(a+l)n.$ ...


1

Naively this seems to test you the Hessian test for local minimum applied to the function $3f+g$. But the test indicates that the Hessian determinant $D(a,b)<0$ (using notation in the link), i.e. $(a,b)$ corresponds to a saddle point. So one needs to modify this approach. The following is a brief sketch: Without loss of generality, assume $p=(a,b)=(0,0)$ ...


1

If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g\circ f$ is continuous at $a$. Since your $g$ is continuous everywhere, it follows that $g \circ f$ is continuous at every point where $f$ is continuous. Said another way, if $g \circ f $ is discontinuous at $a$, then so is $f$. So yes, your conclusion is correct.


4

If you know the answer for the bounded case, then the unbounded case can be easily resolved by considering $g(x) = \arctan(f(x))$ instead. ($f$ is measurable if and only if $g$ is measurable.)


4

Use the hint and multiply the suggested form by your matrix and set it equal to the identity matrix. $$ (cI+dab^T)(I+ab^T) = cI +(c+d)ab^T+da\overbrace{b^Ta}^{\text{scaler}}b^T=cI +(c+d+db^Ta)ab^T=I $$ Since $ab^T$ cannot be a (nonzero) multiple of identity, the only way this sum is equal to identity is that the coefficient $(c+d+db^Ta)$ in front of $ab^T$ ...


1

We have G defined by the properties $$\begin{align} & AGA = A\\ & GAG = G \\ \end{align}$$ Taking the transpose of both sides of each equation $$\begin{align} & (AGA)^T = A^TG^TA^T = A^T\\ & (GAG)^T = G^TA^TG^T = G^T \\ \end{align}$$ gets you the property you want.


0

Without numbers it's theoretically possible and there is no theoretical reason it can't be extrapulated in theory. But we know puppies tend to be very small at birth an go through a rapid weight gain in the first few weeks. But week $20$ the puppy is several times heavier than it was at birth. In fact it's nearly full grown! The growth rate from $20$ ...


2

If your equation is $W=ma+c$, where $W$ is weight, $a$ is age, and $m$ and $c$ are constants, then let $a=0$ to find the weight at birth. Presumably the answer to the question is that your value $c\leq 0$, which gives a weight at birth $\leq 0$, which is impossible.


0

It depends on your assumptions, clearly, the weight of the dog may or may not be directly proportional to its age. Thus, modeling it on a straight line may not be possible if we are to model the actual situation. But, with proper assumption that the weight increase with age. Then, it's plausible.


1

Before studying anything else more advanced than Calculus (at least American Calculus, not sure about other countries) , you must 100% make sure you have a good understanding of basic mathematical logic and basic set theory: "the symbols $\forall$ (for all), $\exists$ (there exists), the logical connectives, and etc. One good place to learn all of this is in ...


2

Linear Algebra I, Analysis I, Probability and Geometry are foundation level courses and can be studied in any order - or indeed, in parallel. Other foundation level topics could be Groups, Topology, Logic and Number Theory. Introductory Calculus will be applying the results of Analysis I (or you can think of Analysis I as being the theory behind Introductory ...


1

For fixed $g$ note that $\sup\limits_{x,y\in T} \langle g,x-y \rangle = \sup\limits_{x,y\in T} |\langle g,x-y \rangle|$, hence $$\left(\sup\limits_{x,y\in T} \langle g,x-y \rangle \right)^2=\left(\sup\limits_{x,y\in T} |\langle g,x-y \rangle|\right)^2=\sup\limits_{x,y\in T} |\langle g,x-y \rangle|^2=\sup\limits_{x,y\in T} \langle g,x-y \rangle^2$$ Let $F:...


2

What you wrote is false. Let $u_i\in A$. Then $z=\sum_{i=1}^r\lambda_i u_i\in Span(A)$. is not correct as each $u_i$ is in $A$ or $B$, but you can’t say that because one of them is in $A$ it is the case for all of them. What you can write is $$\sum_{i=1}^r \lambda_i u_i= \sum_{\substack{1\le i \le n\\u_i \in A}}\lambda_i u_i+\sum_{\substack{1\le i \le n\...


0

Clearly, solution $y=\frac{2}{3}x + \frac{17}{9} $ is satisfying your ODE. May be you are forgetting to substutute y on right hand side.


2

The only guarantee is that if you substitute a (correct) solution for a differential equation back into that equation, you will find that the equation is satisfied. This is the case here. The solution is $y = \frac 23 x + \frac{17}9$ The right hand side of the differential equation is $-2x + 3y - 5 = -2x + 3(\frac 23 x + \frac{17}9) - 5 = \frac 23$. The ...


2

$\int_0^{1} \mu [x,1]dx=\int_0^{1}\int_{[x,1]} 1d\mu (z) dx=\int \int_0^{z}dx d\mu(x)=\int zd\mu(z)$. The second equality follows by Fubini's Theorem.


2

Some properties are immediately established: $\mu(\varnothing) = \lim_{n\to\infty} \mu_n(\varnothing) = 0$. $\mu$ is positive, i.e., for each $E \in \mathcal{F}$, we have $\mu(E) = \lim_{n\to\infty} \mu_n(E) \geq 0$. $\mu$ is additive, i.e., for each disjoint $E_1, \cdots, E_k \in \mathcal{F}$, we have $$ \mu(\cup_{i=1}^{k} E_i) = \lim_{n\to\infty} \mu_n(\...


0

No, uniform convergence for sequences does not imply that the sum of the limit is the limit of the sum. That's true for integrals on finite measure spaces, but the sum of a sequence is the integral wrt counting measure on $\Bbb N$, an infinite measure. To be specific, let $$f_n(j)=\begin{cases}\frac1n,&(1\le j\le n),\\0,&(j>n).\end{cases}$$Then $...


0

At least we can prove the following result. Let $\nu$ be the counting measure on $\mathbb N$ then $$\sum_{k=1}^\infty \mu_n(A_k)=\int_{[1,\infty)} \mu_n(A_k)\,\nu(dk).$$ From Fatou's lemma, we have $$\mu(\cup_k A_k)=\lim_{n\to\infty}\sum_k \mu_n(A_k)\geq\int_{[1,\infty)}\lim_{n\to\infty}\mu_n(A_k)\,\nu(dk)=\int\mu(A_k)\nu(dk)=\sum_k\mu(A_k).$$ So if we are ...


2

You cannot prove this easily from definitions. It is an easy consequence if Vitali -Hahn -Saks Theorem. [ https://en.wikipedia.org/wiki/Vitali%E2%80%93Hahn%E2%80%93Saks_theorem ] This theorem assumes that the measures $\mu_n$ are absolutely continuous w.r.t. some finite measure $m$ but this is always true: we can take $m (E)= \sum\limits_{k=1}^{\infty} \...


1

In my experience, to really remember math for a long time, two things need to happen. 1) You need to really understand WHY a mathematical statement is true, in as many different ways as possible. Can you think of a visual explanation? Can you think of another method to solve the problem? Your goal should be to interact with the material in as many ...


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