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Hot answers tagged random-variables

3

It's enough to show that $E|X|^p$ is continuous in $p$. Why not use dominated convergence: $p\mapsto |X|^p$ is continuous, and if $p\in[1,p_0]$ then $|X|^p\le 1+|X|^{p_0}$. (Think of $p_0>1$ as large but fixed.) In this way you show that $p\mapsto E|X|^p$ is continuous on $[1,p_0]$ for each $p_0>1$.

3

Since the mapping $$(y,p) \mapsto y^{1/p} = \exp( \frac{1}{p} \log y)$$ is clearly continuous, it suffices to show that $$p \mapsto \mathbb{E}(|X|^p)$$ is continuous. Fix $p \geq 1$ and some sequence $p_n \to p$. Choose $N$ sufficiently large such that $p_n \leq N$ for all $n$. Since $$|X|^{p_n} \leq 1_{|X|\leq 1}+ |X|^{p_n} 1_{|X|>1} \leq 1+ |X|^N ... 2 In essence, you have a population of n balls, of which n_1 are red, from which you extract a sample (drawn without repetition) of size r, which will contain a random variable, r_1, of red balls and wish to evaluate \mathsf P(k_1\leq r_1\leq \min\{n_1,r\}, k_2\leq r-r_1\leq \min\{n-n_1, r\}). The count of red balls within the sample, r_1 is a ... 2 The cdf of Y is$$F_Y(y)=P(Y<y)=P(X^3<y)=P(X<\sqrt[3]y)= \begin{cases} 0,&\text{ if } y<0\\ \int_0^{\sqrt[3]y}\frac{x^2}{9} \ dx&\text{ if } 0\le y \le 27 \\ 1&\text{ otherwise} \end{cases}.$$Between 0 and 27$$F_Y(y)=\int_0^{\sqrt[3]y}\frac{x^2}{9} \ dx=\frac1{27}\left[x^3\right]_0^{\sqrt[3]y}=\frac1{27}y.$$It follows ... 1 Given random variables X, Y : \Omega\to\mathbb R defined on the same probability space \Omega, the definition of X+Y is simply the pointwise sum:$$(X+Y)(\omega) = X(\omega)+Y(\omega)$$If X and Y are defined on different probability spaces, X:\Omega_1 \to \mathbb R and Y:\Omega_2 \to \mathbb R, then X+Y is undefined. However, in this case we ... 1 An alternative would to be note that$$ |X|^p \leq (1 + |X|)^p \leq (1+|X|)^{p_0} $$for all 1 \leq p \leq p_0. Note that the right-hand side is an integrable function. Using the dominated convergence theorem, it thus follows that$$ [1, p_0] \to \Bbb{R}, p \mapsto \Bbb{E}|X|^p $$is continuous. Since p_0 >1 is arbitrary, we see that p \mapsto ... 1 The probability sought for is$$p=\lambda_1\lambda_2\lambda_3\iiint_A e^{-(\lambda_1x+\lambda_2y+\lambda_3z)}\ dxdydz$$where$$A=\{(x,y,z):x,y,z\ge 0,x<1,x<y,x<z \}.$$So,$$p=\lambda_1\lambda_2\lambda_3\int_0^1 e^{-\lambda_1x}\left[\int_x^{\infty}e^{-\lambda_2y}\ dy\int_x^{\infty}e^{-\lambda_3z}\ dz\right]\ dx==\lambda_1\int_0^1 ...

1

Let $\bar x = (x_1+\cdots+x_n)/n$ and recall from algebra that $$\sum_{i=1}^n (x_i-\theta)^2 = n(\bar x-\theta)^2 + \sum_{i=1}^n (x_i-\bar x)^2.$$ Then deal with the density: \begin{align} f_{X_1,\ldots,X_n}(x_1,\ldots,x_n) \propto {} & \prod_{i=1}^n\frac 1 \theta \exp\left( \frac{-1} 2 \left( \frac{x_i-\theta}{\theta} \right)^2 \right) = \frac 1 ...

1

You could call $Y$ a generalized multi-variate indicator function that gives values $-1$ or $1$ on the different parts of the support. Note you have to choose which sign to assign to $0$ (however, a single point that is not a point mass doesn't change the values of any integrals). In general there is no closed form for $P(X = v)$. However you can express it ...

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If $X_1,X_2\sim\operatorname{Exp}(\lambda)$ are independent with densities $f$ and $g$ then the density of $X_1+X_2$ can computed as \begin{align} h_2(t) &= (f\star g)(t)\\ &= \int_0^t f(s)g(t-s)\ \mathsf ds\\ &= \int_0^t \lambda e^{-\lambda s}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \lambda^2 e^{-\lambda t} \int_0^t \ \mathsf dt\\ &= ...

1

Consider the sequence of the $k$ trials for a hypergeometric r.v. $X$. As each success occurs, the probability of a success in the next trial is reduced. Similarly, each failure that occurs reduces the probability of subsequent failures. Thus, outcomes that deviate by large amounts from the mean are made less probable compared to the corresponding ...

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Nonetheless, the info that you give, already lets us resolve the problem, if by CDF we mean a CDF of a probability distribution. If a real function $F$ is a CDF, then in particular $0\leq F(x) \leq 1$ for every real $x$ (because $F(x) = P(Z \leq x)$ for some random variable $Z$). Here $F(3) > 1$, hence $F$ cannot be a CDF of a probability distribution.

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The four necessary properties of a CDF are: non decreasing. right continuous. $\lim\limits_{x\to-\infty} F(x) = 0$ $\lim\limits_{x\to+\infty} F(x) = 1$ You have a suspected-to-be cumulative distribution function: $$F(x) = \begin{cases} 0 & : x \leq 0 \\ x^2/3 & : 0\leq x\leq 3 \\ 1& : 3 \leq x\end{cases}$$ Although you've only provided the ...

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First off, an event is a subset of the sample space; it doesn't make sense to talk about a probability distribution for $E$. As for what "prior" means, $T$ is an exponential random variable with unknown parameter $\lambda$. The problem is saying that $\lambda$ itself is a random variable, with $U(a,b)$ distribution. So the probability density function of is ...

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