I have two random variables whose PDF are parameterized by an unknown constant as follows:  P(A;d) P(B;d)  apparently, these two are not independent, so to find P(A+B;d) one cannot use convolution.   1- Do you have any idea how to procede?  2- is my statement about independence correct?  Thanks
121960
I know, I can't use convolution.  I have two random variables A and B and they're dependent. I need Distributive function of A+B
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Let $Z$ be a random uniformly distributed variable on $[0,1]$. Show that the random variables $X = \sin 2\pi Z$ and $Y = \cos 2\pi Z$ are not independent, but nevertheless Cov$[X,Y]=0$.  This is a homework assignment, but I'm a bit stuck.  My thoughts  We can see that $X$ and $Y$ are not independent, since both depend on $Z$. If we want to show this explicitly, then we need to show that $$f_{X,Y}(a,b) \neq f_X(a)\;f_Y(b),$$ where $f_{X,Y}(a,b)$ is the joint probability distribution function. But how can I find the (joint) probability distribution function(s) $f_X, f_Y$ and $f_{X,Y}$?   If I can find these functions, I can also solve the covariance problem. Is this the right way? Or is there a 'better' way to solve this problem?
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Let $X_1$ and $X_2$ be two independent random variables with distribution $U(0,1)$. Let $Y=|X_1-X_2|$. Find the distribution function and density of $Y$.   Solution:  f(x1)=1 whenever x1 between 0 and 1          =0 Else where f(x2)=1 whenever x2 between 0 and 1          =0 Else where f(x1,x2)=1 whenever x1,x2 between 0 and 1           =0 Else where G(y)= P[-y &lt;= x1-x2 &lt;= y]       =\int_{-y}^y \int_{x1_y}^{y+x1} 1 dx2 dx1       =\int_{-y}^y -2y dx1       =4y^2 when y between 0 and 1       =0 when y less than 0       =1 when y greater than 1 Thus, g(y)= \diff G(y) =8y  Is it correct?
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Could anybody help me with solving such problem:  You are given two independent random variables $X$ and $Y$ with continuous uniform distribution. You are to find expected value $E$ of random variable $Z = \max(X,Y)$.
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I have an answer but I'm not sure if I'm doing it right: $f(x)$ is the pdf $f(x)=2e^{-2x}$, then, if the random variable $Y=\ln(X)$, then we simply substitute:  $P(Y \leq y)$=  $P(2e^{-2X} \leq y)$  $P(-4X \leq y)$ since $y=ln(x)$  $P(X \leq - \dfrac {1} {4} y)$  This is the distribution function, right?  Thanks!
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$X_1$ and $X_2$ are two independent random variables whose expected values are known. I am trying to find the expected value of $(X_1/(X_1+X_2))$. Since $X_1$ and $X_1+X_2$ are dependent, I tried to use the formula for covariance to calculate the expected value of the product but this results in variance ($X_1$) being part of the resultant expression which is unknown. Is there any other way of finding the expected value of the expression $(X_1/(X_1+X_2))$?
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Let $X$ and $Y$ be independent $Exp(1)$-distributed random variables. Find the conditional distribution of $X$ given that $X + Y = c$ ($c$ is a positive constant).  this is my idea:  $$f_{X \mid X+Y}(x\mid c) = \frac{f_{X,X+Y}(x,c)}{f_{X+Y}(c) } $$  if this is the right definition then i might calculate the righthand side to obtain the distribution but Iam unsure
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If $X \sim \mathrm{Normal}(\mu,\sigma^2)$ and $Y \sim \mathrm{Normal}(\mu,\sigma^2)$ are independent random variables, how do I prove that $X+Y$ and $X-Y$ are also independent?  What happens with the independence between $X+Y$ and $X-Y$ when $X \sim \mathrm{Normal}(\mu_x,\sigma_x^2)$ and $Y \sim \mathrm{Normal}(\mu_y,\sigma_y^2) $  Thank you
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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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Say, $X$ is dependent on $\alpha$. Rigorously speaking,    if $X$ and $\alpha$ are both random variables, we could write $p(X\mid\alpha)$; however, if $X$ is a random variable and $\alpha$ is a parameter, we have to write $p(X; \alpha)$.   I notice several times that the machine learning community seems to ignore the differences and abuse the terms.  For example, in the famous LDA model, where $\alpha$ is the Dirichlet parameter instead of a random variable.    Shouldn't it be $p(\theta;\alpha)$? I see a lot of people, including the LDA paper's original authors, write it as $p(\theta\mid\alpha)$.
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Trying to find examples of discrete and continuous random variables that are dependent and uncorrelated.    For the continuous I take standard normal $X \sim N(0,1)$ and $Y=X^2$.  By symmetry it is clear that $E(XY) -E(X)E(Y) = E(X^3) - E(X)E(X^2) = 0 - (0)(1) = 0$, so uncorrelated.   But for events $A = \{0 &lt; X &lt; 1\}$ and $B = \{X^2 &gt; 1\}$, I see that $P(A \cap B) = 0$ which does not equal $P(A)P(B)$ since both $A$ and $B$ have nonzero probability. Thus, $X$ and $Y$ are dependent and uncorrelated.  Now I am having trouble constructing a simple example with discrete random variables.
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As asked in the title? Does the independence of two random variables $X$ and $Y$ imply the independence of $X$ and $X+Y$? If so, what's the easiest way to prove that?
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If we know that two events $A$ and $B$ are independent, can we say that $A$ and $B$ are also conditionally independent given an arbitrary event $C$?  $$P(A\cap B) = P(A)P(B) \overset{?}{\Rightarrow} P(A\cap B|C) = P(A|C)P(B|C)$$
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Is there a simple example showing that given $X,Y$ uncorrelated   (covariance is zero), $X,Y$ are not independent?   I have looked up two references, however, I am dissatisfied with both.   In Reference $1$, $X,Y$ are assumed to be independent uniform RVs from $(0,1)$, construct $Z = X+Y, W = X - Y$, then the claim is that $Z,W$ is uncorrelated but not independent. Unfortunately, finding the PDF of $Z,W$ is not trivial. In Reference $2$, $\phi$ is assumed to be uniform RV from $(0,    2\pi)$, and construct $X = \cos(\phi)$, $Y = \sin(\phi)$. Then the claim is that $X,Y$ are uncorrelated but not independent. Unfortunately, the PDFs of $X,Y$ takes on the form of rarely mentioned arcsine distribution.   I just wish to have an example at hand where I can whip out to show that uncorrelated does not necessarily implies independent. Is this do-able?
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Problem: Let $X$ and $Y$ be independent and suppose that each has a $\text{Uniform}(0,1)$ distribution. Let $Z = \min\{X, Y\}$. Find the density $f_Z(z)$ for $Z$. Hint: It might be easier to first find $\mathbb{P}(Z &gt; z)$.  Attempted Solution:  Given that $X, Y \sim \text{Uniform}(0,1)$, how do we not just have the following?  $$ f_Z(t) = f_X(t) = f_Y(t) = \begin{cases} 1 &amp; \text{if } 0 \le t \le 1 \\                                           0 &amp; \text{otherwise}                             \end{cases} $$  Of course I'm highly suspicious of this answer because it's not making use of the fact that $X$ and $Y$ are independent, nor is it making use of the provided hint.
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Let $X_n$ be iid random variables with $P(X_n=(-1)^kk)=\dfrac{C}{k^2\log k}$ for all $k\geq 2$. Here $C$ is just a constant so that the sum of probabilities is $1$. Find constant $\mu$ such that $S_n/n\to \mu$ in probability where $S_n=\sum_{k=1}^nX_k$.   Here $E(X_n)=\infty$. I'm a bit lost after this. In general how does one find constants like this?
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Let $X$ ~ $U(0,2)$ and $Y$ ~ $U(-10,10)$ be two independent random variables with the given distributions. What is the distribution of $V=XY$?   I have tried convolution, knowing that   $$h(v) = \int_{y=-\infty}^{y=+\infty}\frac{1}{y}f_Y(y) f_X\left (\frac{v}{y} \right ) dy$$  We also know that $f_Y(y) = \frac{1}{20}$,    $$h(v)= \frac{1}{20} \int_{y=-10}^{y=10} \frac{1}{y}\cdot \frac{1}{2}dy$$ $$h(v)=\frac{1}{40}\int_{y=-10}^{y=10} \frac{1}{y}dy$$  Something tells me, there is something weird here since it is discontinuous at 0. Please help.