# Tag Info

## Hot answers tagged random-variables

2

Let $N$ be the least integer such that $\sum\limits_{j=1}^{N}\epsilon_j \geq t$. Then $E_i$ is nothing but the set $\{N=i\}$ and these are disjoint. Since $N$ takes then values $1$ to $n$ we see that $\{\max_i \sum\limits_{j=1}^{i}\epsilon_j \geq t\} =\bigcup_i E_i$.

2

Yep, you're almost finished! The hard part was computing that $$\psi_Y(t) = \frac 1 4 + \frac 1 4 e^t + \frac 1 2 e^{2t}; \tag{\ast}$$ and remember that the definition of the moment generating function is $$\psi_Y(t) = \mathbb{E}(e^{ty}) .$$ Now, since we are finding a pmf then we have a discrete random variable and so we appeal to the discrete ...

2

This is generally true for any random variables you wish with finite means. In other words, if $X,Y$ are random variables with $\mathbb{E}[X] < \infty$ and $\mathbb{E}[Y] < \infty$, then $\mathbb{E}[X+Y] = \mathbb{E}[X] + \mathbb{E}[Y]$. In addition, $\mathbb{E}[cX] = c\mathbb{E}[X]$ for any constant $c \in \mathbb{R}$. Together, these are known as the ...

2

Let $F_X(a) = P(X \leq a)$, $F_Y(b) = P(Y \leq b)$ and $F_{X,Y}(a,b) = P(X \leq a \mbox{ and } Y \leq b)$ denote the cumulative distribution functions of $X$, $Y$ and $(X, Y)$, respectively. To show independence, we want to prove that $F_X(a)F_Y(b) = F_{X, Y}(a,b)$ for all $a, b \in \mathbb{R}$. Conditional probability tells us that: $$P(X > a | Y < b) ... 2 For a given random variable, X, the positive part is$$ X^+ = \max\{0,X\} $$and the negative part is$$ X^- = \max\{0,-X\} \text{.} $$Notice that both parts are nonnegative. This confirms the comment to the Question: "???" is 0. This definition exactly parallels the definition of positive and negative parts of a function. 2 Let's say we had some random variables X and Y such that P(X > a \cap Y > b) = \frac{a}{b}P(X > a)P(Y > b) for b > a > 0, else P(X > a)P(Y > b). Clearly, letting a = b would satisfy the first criterion:$$P(X > a | Y > a) = \frac{P(X > a \cap Y > a)}{P(Y > a)} = \frac{\frac{a}{a}P(X>a)P(Y>a)}{P(Y>a)}...

1

If $X$ and $Y$ are independent standard normal random variables, then using polar coordinates you can write (for any arbitrary bounded measurable function $g$) $$E[g(X,Y,\sqrt{X^2+Y^2})] =\int_0^{2\pi} \int_0^\infty g(r\cos\theta,r\sin\theta,r) r {e^{-r^2/2}} dr{1\over 2\pi} d\theta.$$ This shows that the joint distribution of $(A,B,Z)$ is the same as that ...

1

$f_X(x) = \lambda_1 e^{-\lambda_1 x}, f_Y(y) = \lambda_2 e^{-\lambda_2 y}$ $f_{XY} (x, y) = \lambda_1 \lambda_2 e^{-(\lambda_1 x + \lambda_2 y)}$ $U = X + Y, V = X-Y$ and using, $x = \frac{u+v}{2}, y = \frac{u-v}{2}$ ...(i) Joint pdf of $U$ and $V$, $f_{UV}(u,v)=\frac{1}{2}\lambda_1\lambda_2 e^{-\frac{1}{2} \big((\lambda_1 +\lambda_2)u + (\lambda_1 - \... 1 If you don't even know something about a single RV, it's impossible to know any more about what happens when you add multiple RVs together. A random variable could: Be discrete or continuous Have finite or infinite support Have finite or undefined mean Have finite or infinite variance Be well-behaved or pathological in some fashion A pair of random ... 1 But I don't know how to find P[B]. Any ideas? Yes, as usual, by definition $$\mathbb{P}[XY>80]=\mathbb{P}\left[Y>\frac{80}{X}\right]$$ thus all you have to do is to integrate$f(x,y)$in the specified region... that is the purple area here below 1 Yes, you are correct. Variables can take on different values for repeats of the same situation. A single random sample is the values of a set of "draws" from a population. In probability we reason from a distribution to data; in statistics we reason from data to a distribution or model. We want to know about the full population, but without a ... 1 For a scalar random varible$z$that has$\mathbb{E}(z) = \bar{z}$, where$\mathbb{E}$means the expectation (mean), by definition of the variance,$Var(z) = \mathbb{E}[(z - \bar{z})^2] = \mathbb{E}[z^2 - 2z\bar{z} + \bar{z}^2] = \mathbb{E}[z^2] - \bar{z}^2$since the expectation is a linear operator. Now, plug in$z = aX + bY + c. \begin{aligned} Var(aX+... 1 The main thing to remeber here is that the probability density function of a uniform distribution isf_X(x) = \frac{1}{A-B} = \frac{1}{1} = 1$$You might immediately ask "doesn't that mean that every point is certain?" and the answer is no because if you integrate around successively smaller intervals around the point you'll converge to 0. Okay, ... 1 This is due to the fact that the support of the uniform joint distribution is [0;1]\times [0;1] Instead of using directly the convolution you can use the jacobian thus you can appreciate the underlying logic. to do that, set$$\begin{cases} z=x+y \\ u=x \end{cases}\rightarrow \begin{cases} x=u \\ y=z-u \end{cases}$$the jacobian is evidently |J|=1 thus$$... 1 Note that\epsilon_i$and$x_i$can be treated as constants, therefore $$E[\epsilon_i | x_i]=E[\epsilon_i]=\epsilon_i$$ Assuming that$\epsilon_i=0$and$P(X=x_i)\gt 0$, then $$E[\epsilon | X]=\sum_{i=1}^n \epsilon_i P(\epsilon=\epsilon_i\vert X=x_i)=0$$ 1 Let$P_n(x)\ (0\leq x<1)$be the probability density for$x=Y$after n iterations. Since we add uniformly random numbers, the probability at some$x$is the integral of the probability in the previous range$[x-1,x]$, namely,$[0,x]$. $$P_0(x)=1,\ P_{n+1}(x)=\int_0^x P_n(t)\,dt$$ Recursively,$\$P_{1}(x)=x,\ P_{2}(x)=\int_0^xP_1(t)\,dt=\frac{x^2}{2},\ \...

1

As a bit of graphical intuition: Let's say you have two independent datasets, like this: %pylab x = randn(10000) + randint(0, 2, 10000) * 5 y = randn(10000) + randint(0, 2, 10000) * 5 fig, axes = plt.subplots( 2, 2, sharex='col', sharey='row', gridspec_kw={'width_ratios': [1, 0.1], 'height_ratios': [0.1, 1]}) axes[1, 0].scatter(x, y, s=1) axes[0, 0]...

Only top voted, non community-wiki answers of a minimum length are eligible