Is it true that   $$\int_{\Bbb R} G(x)dF(x) + \int_{\Bbb R} F(y)dG(y) = 1$$  where $F$ and $G$ are distribution functions of $X$ and $Y$ such that both have no common jump points i.e. $P(X=Y) = 0$ ? How to prove it ? I am unfamiliar with Stieljes integral
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Suppose that $X$ and $Y$ are random variables with distribution functions $F$ and $G$. If $F$ and $G$ have no common jumps then I need to show that $E[F(Y)] + E[G(X)] = 1$.  How to proceed here ? Shall I try to show that the LHS is some probability of almost sure events or product of two almost sure events. Clearly, $F(Y)$ and $G(X)$ are random variables with range in $[0,1]$.
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Let $f,g,g´$ be continous on $[a,b]$ and $g$ monotone on $[a,b]$; then there exist $c\in (a,b)$ so that $$\int_{a}^{b}f(x)g(x)dx=g(a)\int_{a}^{c}f(x)dx+g(b)\int_{c}^{b}f(x)dx$$  Ineed to apply the first generalized mean value theorem for integrals i.e. $$\int_{a}^{b}f(x)g(x)dx=f(c)\int_{a}^{b}g(x)dx$$ to $g´$and $\int_{a}^{x}f(x)dx$ but I don´t know how please I would appreciate your help
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I have two independent variables: $X$ follows from standard Gaussian distribution $N(0,\sigma^2)$; $Y$ follows from Rademacher distribution, i.e., $Y$ can be either $-1$ or $1$ with the same probability.  My question is:  For a new random variable of $Z=XY$, What is the probability density function?   Or can we have some concentration inequalities for $Z$ (e.g., show how $Z$ approaches its expectation, etc.)?
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Suppose $f: D \to \mathbb R$ and $g:D \to \mathbb R$, $x_0$ is an accumulation point of $D$, and $f$ and $g$ have limits at $x_0$. If $f(x) \le g(x)$ for all $x \in D$, then   $\lim \limits_{x \to x_0}f(x) \le \lim \limits_{x \to x_0}g(x)$  Assume that $\forall x \in D$,   $f(x)$ and $g(x)$ both have limits at some point $x_0$ and $f(x) \le g(x)$    We know that $\lim \limits_{x \to x_0}f(x) = f(x_0)$ and $\lim \limits_{x \to x_0}g(x)=g(x_0)$  $\therefore f(x_0) \le g(x_0)$ because $f(x) \le g(x) $  $\forall x \in D$
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$X$ and $Y$ are independent and identically distribued (i.i.d.), $X+Y$ and $X-Y$ are independent, $\mathbb{E}(X)=0$ and $\mathbb{E}(X^2)=1$. Show that $X\sim N(0,1)$.  We should use characteristic functions to prove this. Any ideas?
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I'm solving a problem where I've this 'expectation': $$ \int_{0}^y x\cdot f(x) dx $$ where $f(x)$ is a PDF with support on $[0, z]$, with $z&gt;y$. Is there a way to rewrite it without the integral and as a function of the CDF? I've tried integration by parts, but without great success: $$ \int_{0}^y x\cdot f(x) dx = y\cdot F(y) -\int_0^y F(x) dx $$ I have hard time to solve the second part.
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Let's say - $F(x) &gt; G(x)$ for all $x$ in $E$.  So is it true that ?  $$\int F(x)\,dx  &gt; \int G(x)\,dx$$ or I can use this statement only when the two functions are non-negative functions.  Thanks!
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How to show $f=g$ almost everywhere  Let two real-valued measurable functions $f$ and $g$ be such that for any measurable set $E$ the integrals $\int_E f\,d\mu$  and  $\int_E g\,d\mu$ coincide.  Could you please help.  It seems obvious. we could define the integrals as charges. can we do sth from here. but I do not know how to approach.
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Show that for a non-negative random variable $X$,   $$\mathbb E(X)=\int_0^\infty \mathbb P(X\ge x) \, dx.$$   I started with  $$\mathbb E(X)=\int_0^\infty x \, dF(x)=\int_0^\infty \int_0^x dt\,dF(x).$$ This is my try.
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I was wondering if there is a name for a function that satisfies the conditions  $f:\mathbb{R} \to \mathbb{R}$ and $f(x+y)=f(x) \cdot f(y)$?  Thanks and regards!
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Let $(B_t)_{t\geq 0}$ be a standard Brownian motion in $\mathbb{R}^d$. It is intuitive that, for fixed $s&lt;t&lt;u$  $$\mathbb{E}[B_t\mid \sigma(B_s,B_u)]=B_s+\frac{t-s}{u-s}(B_u-B_s).$$  However, I cannot think of a way to show this rigorously.   If first attempted to take $A\in\sigma(B_s,B_u)$ and show that $\mathbb{E}[1_A B_t]=\mathbb{E}[1_A(B_s+\frac{t-s}{u-s}(B_u-B_s))]$. But I cannot manage to show this equality.   I'd be very thankful for any ideas and suggestions on how to tackle this problem.
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Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated.
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This question is a possibly harder version of: .  Question.  Let $f,g :\mathbb R\to\mathbb R$, such that \begin{align} f(x-y)=f(x)\, g(y)-f(y)\, g(x), \tag{1}\\ g(x-y)=g(x)\, g(y)+f(x)\, f(y), \tag{2} \end{align} for all $x,y \in \mathbb{R} $. If $g$ is continuous at $x=0$ and not identically zero, then there exists an $\alpha\in\mathbb R$, such  $$ f(x)=\sin \alpha x\quad\text{and}\quad g(x)=\cos \alpha x. $$ Is there a pair of discontinuous $f$ and $g$ satisfying $(1)$ and $(2)$?  Update. If $\ell :\mathbb R\to\mathbb R$ is a linear functional over $\mathbb Q$ (i.e., $\ell(qx+ry)=q\ell(x)+r\ell(y)$, for all $x,y\in\mathbb R$ and $q,r\in\mathbb Q$), then $$ \sin\big(\ell(x)\big), \quad \cos\big(\ell(x)\big), $$ satisfy $(1)$ and $(2)$. Discontinuous such functionals do exist, and they are obtainable using Zorn's Lemma (equivalently the Axiom of Choice.) This takes care of the second question.
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Written with .      Let $f: \Bbb R \to \Bbb R $ be a continuous function such that $f(i) = 0 \ for \ all \ i \in \Bbb Z$. Then which of the following is true : -          A. Image(f) is closed in $\Bbb R$     B. Image(f) is open in $\Bbb R$     C. f is uniformly continuous     D. None of the above   For the correct option, look at the end of the question.     I tried to contradict the options by bringing examples(It was obviously the first step considering 'None of the above was' an option)      I wasn't able to think of a single function that contradicts either of the options. However, graphically I would say that these functions satisfy the hypothesis and contradict the options (Image uploaded)  Here's what the functions are   1) Just like $sin x,$ the function touches 1 and -1. The major difference would be that this function touches the x-axis at all integral values and no where else.  2) The second graph touches the integers and zero and in each interval $(m,m+1)$, the graph's maximum and minimum value tend to increase slightly and they tend to +1 and -1 as $x \to \infty$ Thus, the function gets arbitrarily close to 1 and -1 without touching them.    3) The third graph is just like $x Sin(x)$. The difference is that this one touches the x-axis at integer values. With each interval $(m,m+1)$, our graph has its 'magnitude of difference' increasing.    Are my graphs correct? If they, can they explicitly be expressed as familiar functions? If not, which explicitly express-able  examples can be used to contradict the first three options?  The correct answer is D    Source - Tata Institute of Fundamental Research Graduate Studies Exam 2016
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Let , $f:\Bbb C \to \Bbb C$ be an entire function and $\displaystyle \int_{\Bbb R^2}|f(x+iy)|\,dx\,dy&lt;\infty$. Then prove that $f(z)=0$ for all $z\in \Bbb C$.   Put $x=r\cos \theta$ and $y=r\sin \theta$. Then , integration becomes $\displaystyle \int_{r=0}^{\infty}\int_{\theta=0}^{2\pi}|f(re^{i\theta})|r\,d\theta\,dr&lt;\infty$. But from here I'm unable to conclude nothing.! How I proceed? Any hint please.
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So assume $Y$ and $X$ are exponentially distributed with parameters $y_1$, and $x_1$ respecitively.  assume c is a constant.  I am having huge trouble to understand the integration of the following expression.  $P(Y&lt;c/u(X))$  $=\int_{t}^{\infty}f_X(x)\int_{0}^{c/u(x)}f_Y(y)dydx +\int_{0}^{t}f_X(x)\int_{c/u(x)}^{\infty}f_Y(y)dydx$  where t is the cross-point that $u(x)$ change sign    here $c/u(x)$ is given by the plot below, t is the point crossing the zero:   Confusion:   I don't understand the second integration of the second term "$\int_{c/u(x)}^{\infty}f_Y(y)dydx$", this isn't right because Y an X only defined for y>0, and x>0. So it's the first quadrant in this plot.
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If $f(x)$ is a density function and $F(x)$ is a distribution function of a random variable $X$ then I understand that the expectation of x is often written as:   $$E(X) = \int x f(x) dx$$  where the bounds of integration are implicitly $-\infty$ and $\infty$. The idea of multiplying x by the probability of x and summing makes sense in the discrete case, and it's easy to see how it generalises to the continuous case.  However, in Larry Wasserman's book All of Statistics he writes the expectation as follows:  $$E(X) = \int x dF(x)$$  I guess my calculus is a bit rusty, in that I'm not that familiar with the idea of integrating over functions of $x$ rather than just $x$.   What does it mean to integrate over the distribution function? Is there an analogous process to repeated summing in the discrete case? Is there a visual analogy?   UPDATE: I just found the following extract from Wasserman's book (p.47):     The notation $\int x d F(x)$ deserves some comment. We use it merely   as a convenient unifying notation so that we don't have to write   $\sum_x x f(x)$ for discrete random variables and $\int x f(x) dx$ for   continuous random variables, but you should be aware that $\int x d F(x)$  has a precise meaning that is discussed in a real analysis   course.   Thus, I would be interested in any insights that could be shared about what is the precise meaning that would be discussed in a real analysis course?