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If $F$ is a CDF and it must be shown that $F$ is the CDF corresponding with the uniform distribution on $[0,1]$ then it is enough to prove that $F(y)=y$ for every $y\in(0,1)$. This because the values of $F$ on elements not in $(0,1)$ are determined by the following rules for a CDF: If $y\geq1$ then $1\geq F(y)\geq F(z)=z$ for every $z\in(0,1)$ and ...

4

If $n$ is the number of tosses, then $T=n-H$, so $H-T=2H-n$. Since $n$ is large, $H$ has a close to normal distribution, mean $n/2$, and variance $(n)(1/2)(1/2)$. So $2H-n$ has close to normal distribution, mean $0$ and variance $n$. Let $W$ be a normal with mean $0$ and variance $n$. Let us find the mean of $|W|$. This is $$\int_{-\infty}^\infty \frac{|w|}... 3 Your last line should have been$$m_{X+Y}(t) = \int_{-\infty}^\infty e^{ts} f_{X+Y}(s) \mathop{ds}= \int_{-\infty}^\infty e^{ts} \int_{-\infty}^\infty f_X(s-y) f_Y(y) \mathop{dy} \mathop{ds}.Making the change of variables s=x+y gives you the answer. 3 tl;dr: look at the characteristic function, invoke independence of the summands to get a product of characteristic functions, compute the closed-form expression, and finally squint hard at the resulting expression to recognize a known characteristic function. (This is a good method whenever you have a r.v. defined as the sum of independent (possibly ... 3 According to Earliest Known Uses of Some of the Words of Mathematics, the term "characteristic function" in this sense (actually its French equivalent "fonction caractÃ©ristique") was first used by PoincarÃ© in 1912 (except that with his notation, that function was what we now call the moment generating function). To me as an analyst, the terminology never ... 2 We can calculate the expected number of 1's using the linearity of expectation. Let X_1 denote the random variable that is 1 if the 4-sided die rolls a 1, and 0 otherwise. Similarly, let X_2 denote the indicator variable for the 6-sided die, and X_3,\ldots, X_7 for the other dice. We have X_1+X_2+\cdots +X_7 indicating the total number of 1's ... 2 Based on your assumption, X and Y are jointly normal. Let \rho be the correlation, that is, \begin{align*} \rho = \frac{E\big((X-E(X))(Y-E(Y)) \big)}{\sqrt{E\left((X-E(X))^2 \right)}\sqrt{E\left((Y-E(Y))^2 \right)}}. \end{align*} Then, Y-E(Y) and \begin{align*} Z:=\frac{X-E(X)}{\sqrt{E\left((X-E(X))^2 \right)}} -\rho \frac{Y-E(Y)}{\sqrt{E\left((Y-E(Y)... 2 These questions make little sense from the classical (Kolmogorov's axiomatic) probability theory point of view. For example, let \xi be U[0,1], and \eta = \xi if \xi is irrational and 1/\pi otherwise. Then these variables have the same distribution, yet \eta never takes rational values. So the answers to both questions would be 0. However, ... 2 I will try to start from the simplest case possible and then build up to your situation, in order to hopefully develop some intuition for the notion of convolution. Convolution essentially generalizes the process of calculating the coefficients of the product of two polynomials. See for example here: Multiplying polynomial coefficients. This also comes up ... 2 You could use a summation, as noted in the comments. Or we can notice that P(X> x) = P(X\geq x+1). In words, this means we have fail the first x trails; success has to happen on the x+1 trial, or the x+2 trial, etc. The probability of failing on any particular trial is1-P(\text{double ones}) = 1-\frac 16\frac 16 =\frac{35}{36}.$$Hence$$P(X\...

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We start with the Pearson differential equation: $$f'(x)=-\frac{\left(a_1 x+a_0\right) }{b_2 x^2+b_1 x+b_0}f(x),$$ Define $g(x)=b_2 x^2+b_1 x+b_0$ (using a trick in Diaconis et al.(1991)). Consider (f g)'(x), also written $(f(x) g(x))'=f'(x) g(x) +f(x) g'(x)$. We have $$(f g)'(x)=(-a_0 + b_1 - a_1 x + 2 b_2 x) f(x)$$ We assume that the distribution has ...

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This is ugly, but may show how, even having no clue what one is looking for, it's still possible to try something (and possibly arrive at the answer, albeit not in the most elegant way). The important thing being: if you have no clever idea, do the non-clever systematic cumbersome thing. It might work. I don't like numbers, so instead of $2$ I used $a$. The ...

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$VarX_n=EX_n^2=n^{2s},VarS_n=\sum_{j=1}^n j^{2s}$ so chebychev's inequality gives $P(|S_n/n|>\epsilon)\le \sum_{j=1}^nj^{2s}/(n^2\epsilon^2)\sim \int_1^nx^{2s}dx/(n^2\epsilon^2) \sim n^{2s+1}/(n^2\epsilon^2)\to 0$ iff $2s-1<0, s<1/2$. I assumed $s\neq-1/2$ for the integral approximation, but that case is analogous.

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Since $\bar{F}(x) = 1- \mathbb{P}(X \leq x) = \mathbb{P}(X>x)$, we have $$x \bar{F}(x) = \int_{\{X>x\}} x \, d\mathbb{P} \leq \int_{\{X>x\}} X \, d\mathbb{P}$$ for any $x>0$. If $\mathbb{E}(X)<\infty$, then we can let $x \to \infty$ using the dominated convergence theorem to conclude $$\lim_{x \to \infty} x \bar{F}(x)=0.$$ If $\mathbb{E}(X)... 1 There are a couple of slightly different ways you can think about this. Firstly, it's helpful to rewrite the$f_X(x)$term in the denominator of your conditional density as$\int_{\mathbb{R}} f_{X,Y}(x, y) dy$- that is, the integral over the joint density of the event you're conditioning on. By a similar token, the conditional density for$X$conditional ... 1$Y=X^2+Z$means$Z=Y-X^2$, so we let$z(x,y)=y-x^2$After affirming that there is a bijection between$(X,Z)\leftrightarrow(X,Y), (because why?), we then can directly apply the Jacobian change of variables transformation: \begin{align}f_{X,Y}(x, y) =&~ f_{X,Z}(x, z(x,y))~\big/\Big\lvert\dfrac{\partial\big(x,z(x,y)\big)}{\partial\big(x,y\big)}\Big\... 1 Your criterion for positive definiteness is incorrect. A positive determinant is necessary but not sufficient for positive definiteness (Sylvester's criterion for positive definiteness of a symmetric matrix requires all leading principal minors to be positive.) At the point (0,0) the Hessian matrix for the bivariate standard normal is H:=\left(\begin{... 1 Outline: Draw the 10\times 10 square with corners (0,0), (10,0), (10,10), (0,10). Draw the two lines y=x+2 and y=x-2. We want the probability of falling in the part K of the square that is between these two lines. This is the area of K divided by 100. It is easier to find first the area of the part of the square that is not in K. ... 1(h-1)h'=2 $$is a separable differential equation: it can be simply re-written as$$ \frac{d}{dt}\left(\frac{h^2}{2}-h\right) = 2 $$from which:$$ (h-1)^2=(4t+C) $$and$$ h = 1\pm\sqrt{C+4t} $$readily follow. Have also a look at the generating function for Catalan numbers. 1 The easy way for this problem, as is the case for many pdf problems, is to work with CDF's instead. Here, since f(x) = \frac{x-1}{2} on (1,3),$$ F(x) = \left\{ \array{0 & x\leq1\\ \frac{(x-1)^2}{4} & 1< x < 3 \\ 1 & x\geq 3 }\right. $$And this needs to match the CDF of the uniform distribution ojn (0,1)$$F(y) = y$$So$$y= \frac{... 1 A variant of this problem comes up a lot when you're trying to simulate something using a Monte Carlo code. Here's how I would obtainu(x)$, and it doesn't require the solution of any differential equations: The CDF$F(x)$is given by $$F(x) = \int_1^x f(w) dw = \frac{(x-1)^2}{4}$$$F(x)$is uniformly distributed between 0 and 1 (my wording here may ... 1 The two word answer is "polar coordinates". In more detail, let$f:S^{n-1}\to\Bbb Rbe a continuous function. Then \eqalign{ \Bbb E[f(X)] &=\int_{\Bbb R^n}f(x_1/z,\ldots,x_n/z)(2\pi)^{-n/2}e^{-z^2/2}\,dx_1\cdots dx_n\cr &=(2\pi)^{-n/2}\int_0^\infty\left[\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du)\right]e^{-r^2/2}r^{n-1}\,dr\cr &=c_n\int_{S^{n-1}} ... 1 HINT If X is a random variable with pdf f(x) then \mathbb{E}[g(X)] = \int_{-\infty}^\infty g(x) f(x) dx.$Can you apply this to your problem? What is$g(x)\$? Can you integrate?

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