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0

You can do it by integration. \begin{align} E[\max\{X, 1-X\}] & = \int_0^{1}\max\{x, 1-x\}\operatorname d x \\ & = \int_0^{1/2} 1-x\operatorname d x + \int_{1/2}^1 x\operatorname d x \\ & = \big[x-x^2/2\big]_{x=0}^{x=1/2} + \big[x^2/2\big]_{x=1/2}^{x=1} \\ & = 3/4 \end{align} Alternatively: The break occurs uniformly, half the time it ...

1

The length of the longer piece is $\max\{X,1-X\}$. So your expectation would be $\int_0^1 \max\{x,1-x\}\,dx$.

0

One approach is to find the distribution of $X$. If $0<x<1$ then $$F_X(x) = \mathbb P(X\leqslant x) = \mathbb P(U^2 \leqslant x) = \mathbb P(U \leqslant \sqrt x)=\sqrt x.$$ So the density is obtained by differentiating: $$f(x) = \frac{\mathsf d}{\mathsf dx} F(x) = \frac12 x^{-\frac12},\quad 0<x<1.$$ Now we compute \begin{align} \mathbb E[Y] ...

1

If $f(x)$ is the density of random variable $X$, then $E[g(X)]=\int_{-\infty}^\infty g(x)f(x)dx$. In your case, $Y=g(U)=\exp(U^2)$. So $$E[Y]=\int_{0}^1 \exp(U^2)\cdot 1\cdot du$$

1

$$D = b^2-4a = (4k)^2 - 4 \cdot 4 \cdot (k+2) = 16(k^2 - k - 2) = 16(k-2)(k+1)$$ We need $D \ge 0$ for real roots, which happens when $k \ge 2$ or $k \le -1$. Can you take it from here?

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Hint: \begin{align} E[\min(\theta, I)] &= \int_0^1 \min(t, I) \; \mathrm d t \\ &= \int_0^I \min(t, I) \; \mathrm d t + \int_I^1 \min(t, I) \; \mathrm d t\\ &= \int_0^I t \;dt + \int_I^1 I \;\mathrm d t \\ &= \frac12 I^2 + (1-I)I \\ &= I - \frac12 I^2. \end{align}

0

You're on the right track. The type I error is the probability to reject $H_0$ under $H_0$, so $\alpha=P(X>0.9|\theta = 1)$. Under $H_0$, $X\sim U(\lbrack 0,1\rbrack )$, so you just have to determine the probability that a $U(\lbrack 0,1\rbrack )$ distributed random variable to be greater than $0.9$. The type II is the probability not to reject $H_0$ ...

0

You can indeed do this graphically. Consider A=0 and B=1 for simplicity. Your situation can be represented by this : So the probability you want is the area of the colored area: $$P( X+Y > C) = 1-\frac{C^2}{2}$$ Edit : with the correct graph it's better

-1

that is because of below : ! :D $f_G(x)=\frac{d}{dx} P_r(G(x)<x)$ $f_G(x) = \frac{d}{dx}F_X(x^\frac{1}{3}) =\frac{1}{3}x^\frac{-2}{3}$

0

Since $X\sim \operatorname{Uniform}(0,1)$ we know that $F_X(x) = \mathbb{P}(X \leq x) = x$. Let $Y = X^3$. $$F_Y(y) = \mathbb{P}(X^3 \leq y) = \mathbb{P} (X \leq y^{1 \over 3}) = y^{1 \over 3}.$$ Taking the derivative of the cumulative distribution function we obtain the pdf: $$F'_Y(y) = f_Y(y)= {1\over 3} y^{- {2\over 3}}.$$ We can see from the pdf that $Y$ ...

1

It is true that $f(x)=x^3$ pairs off each member of $[0,1]$ with exactly one other member of $[0,1]$, i.e. this function is a bijection from $[0,1]$ to itself. But continuous probability distributions don't really care about this. (Discrete distributions do care about this.) Instead, if $X$ is uniform on $[0,1]$, then $X^3$ is more likely to be small than it ...

0

Because you have that $$P( X < \frac{1}{2} ) = \frac{1}{2}$$ and $$P( X^3 < \frac{1}{2} ) = P( X < \frac{1}{\sqrt[3]{2} } ) = \frac{1}{\sqrt[3]{2} }$$ And both are clearly different. So the repartition functions are differents. But the density is the anti-derivative of the repartition function, so these are different too.

1

If $f$ denotes the probability of some random variable then this does not mean that $f(x)=P(X=x)$ for each $x$. It seems that you are thinking this way. Yes. In the context of your question: if $y=x^3$ then we have $P(X^3=y)=P(X=x)=0$. But that is not a determining relation if it comes to the PDF of $X^3$.

0

For a random variable $X$ distributed according to the pdf $p_X$ and a differentiable function $f$ with positive derivative, you can show that the random variable $f(X)$ is distributed according to $$p_{f(X)}(y) = \frac{p_X(f^{-1}(y))}{f^\prime(f^{-1}(y))}.$$ In your case, you have that $p_X(x) = \chi_{[0, 1]}(x)$ (with $\chi_{A}$ the ...

0

Hint: If $X$ is uniformly distributed and $a,b\in\mathbb R$ with $a\neq0$ then $Y:=aX+b$ is also uniformly distributed. This in the meaning that its PDF has a positive constant value on a certain set and takes value $0$ on the complement of that set. If you know that the PDF of $Y$ takes constant positive value $c$ on e.g. some interval $(p,q)$ and takes ...

0

No: The cdf of $2X-4$ is $$F_{2X-4}(x)=P(2X-4<x)=P\left(X<\frac{x+4}2\right)= \begin{cases} 0,& \text{ if } x<-4\\ \frac{x+4}2,& \text{ if } -4\le x\le -2\\ 1,&\text{ if } x>-2 \end{cases}.$$ The density is $$f_{2X-4}(x)=\frac{dF_{2X-4}(x)}{dx}= \begin{cases} \frac 12,& \text{ if } -4\le x\le -2\\ 0,&\text{ otherwise } ... 1 You can (approximately) do it using Rejection sampling. 1 Using NIST Standard 800-22, "A Statistical Test Suite for Random and Pseudorandom Number Generators for Cryptographic Applications", I can use Hamming Weight as the basis for a spectral test. That is good enough to expose the bug. 2 I suggest that you check "The Art of Computer Programming", Vol. 2 (Seminumerical Algorithms), by Donald E. Knuth. Chapter 3 is completely devoted to random number generators, and tests to determine randomness. You will find many tests that you can apply to your data. I would recommend the Fourier test, which has yielded good results for me. Nevertheless, I ... 0 The approach through the cumulative distribution function is reasonable, but there are troubles with the algebra. I am interpreting \log as the natural logarithm. Minor modification will take care of things if we interpret \log as logarithm to the base 10. We have for suitable y$$\Pr(Y\le y)=\Pr\left(-\frac{\log(1-X)}{\lambda}\le ...

1

When someone says that a baby is born every (instert a period of time), they actually mean that this is true on average. It can happen in a small village that three babies are born on the same day. Many of these deviations from the average get washed out if you have enough data. But some things don't. Several things affect when people are born. A mother's ...

4

The cited statistic, "every second 3 babies are born" is just a rough average. In truth, human birth rates are extremely seasonal (the weather having a dramatic effect on human behavior). Here is a reference: http://www.sas.upenn.edu/~valeggia/pdf%20papers/birth_seasonality.pdf

0

There are two ways you could tackle the problem: (a) Use "change of variables" from $X$ and $V$ to $X$ and $Y$ and then find the required conditional distribution and expectation. (b) Use the "direct" method, which is what you've done. For this particular problem, I think (a) is a slightly simpler but I will use (b) here since it is in line with your ...

0

If $U\sim U(0,1)$ then $f_U(u)=1 I_{[0,1]}(u)$ and $V\sim\Gamma(2,1)$ then $f_V(v)=ve^{-v}I_{[0,\infty]}(v)$ Now take $W=g_1(U,V)=UV$ and $Z=g_2(U,V)=V$ thus $$J=\begin{bmatrix}\frac{\partial W}{\partial U}&&\frac{\partial W}{\partial V}\\\frac{\partial Z}{\partial U}&&\frac{\partial W}{\partial V}\end{bmatrix}=v\Rightarrow ... 2 Brute force: let S = UV. Then$$\begin{align*} \Pr[S \le s] &= \Pr[UV \le s] = \Pr[V \le s/U] = \int_{u=0}^1 \Pr[V \le s/u] f_U(u) \, du \\ &= \int_{u=0}^1 \int_{v = 0}^{s/u} \frac{b^a v^{a-1} e^{-bv}}{\Gamma(a)} \, dv \, du \\ &= \frac{b^a}{\Gamma(a)} \left( \int_{v=s}^{\infty} \int_{u=0}^{s/v} v^{a-1} e^{-bv} \, du \, dv + \int_{v=0}^s ...

1

You have the right joint density function provided you get the right boundaries: $$f_{S,T}(s,t) = \begin{cases} e^{-t} & \text{if }0<s<t, \\ 0 & \text{otherwise.} \end{cases}$$ (You can draw a picture and see that this is half of the first quadrant.) Now suppose we want the marginal distributions of $S$ and $T$.  f_S(s) = \int_s^\infty ...

0

I am replying to Miller Zhu's comment (2nd comment to question), but the comment box doesn't have enough space. I'm only using the inscribed ellipsoid to get a sense of the principal axes of the polytope. Since you don't have access to the vertices of the polytope, computing the inscribed ellipsoid is the only computationally tractable problem (the ...

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