It is well known that the probability density function characterizes the independence of random variables in the following sense.  $$X,Y \quad\text{independent}\iff f(x,y)=f_x(x)f_y(y)$$  where $f$ is the pdf of the random variable $(X,Y)$ and $f_x$ and $f_y$ are the pdfs of the marginal distributions of $X$ and $Y$, respectively. My question is, seeing as there is a bijection between characteristic functions and probability distributions, do characteristic functions also characterize independence? In the following sense  $$\varphi_{(X,Y)}(t_1,t_2)=\varphi_X(t_1)\varphi_Y(t_2)$$  where again $\varphi$ is the characteristic function of the random variable in its subscript.
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This is solely a reference request. I have heard a few versions of the following theorem:     If the joint moment generating function $\mathbb{E}[e^{uX+vY}] = \mathbb{E}[e^{uX}]\mathbb{E}[e^{vY}]$ whenever the expectations are finite, then $X,Y$ are independent.   And there is a similar version for characteristic functions. Could anyone provide me a serious reference which proves one or both of these theorems?
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$(\Omega, \mathcal{F}, P)$ is a probability space. $X$ is a r.v. defined on it, and $\mathcal{G}_1, \mathcal{G}_2$ are two $\sigma$-algebra, can we claim the following: $$ \mathbb{E}\{\mathbb{E}[X|\mathcal{G}_1]|\mathcal{G}_2\}=\mathbb{E}\{\mathbb{E}[X|\mathcal{G}_2]|\mathcal{G}_1\}=\mathbb{E}[X|\mathcal{G}_1\cap\mathcal{G}_2]. $$ If this is not true, why the following is true: $X_1,X_2,\ldots,X_n$ are independent r.v. on the same probability space, $Z=f(X_1,X_2,\ldots,X_n)$ is another r.v. And we have: $$ \mathbb{E}\{\mathbb{E}[Z|X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n]|X_1,\ldots,X_i\}=\mathbb{E}[Z|X_1,\ldots,X_{i-1}]. $$
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Definition: The c.d.f. $F$ of a random variable $X$ is a function defined for each real number $x$ as follows:$$F(x)=\Pr(X\leq x) \text{ for } -\infty&lt;x&lt;\infty$$    Let $$F(x^-)=\lim_{y\rightarrow x,\,y&lt;x}F(y)$$ and $$F(x^+)=\lim_{y\rightarrow x,\,y&gt;x}F(y)$$       Property of cumulative distribution function: A c.d.f. is always continuous from the right; that is , $F(x)=F(x^+)$ at every point $x$.        Proof: Let $y_1&gt;y_2&gt;\dots$ be a sequence of numbers that are decreasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the intersection of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty} \Pr(X\leq y_n)=F(x^+).$$     Now I think the left inequality can also be proved in the similar way as:  Let $y_1&lt;y_2&lt;\dots$ be a sequence of numbers that are increasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the union of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty}\Pr(X\leq y_n)=F(x^-).$$  Where am I wrong?
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Is there any counterexample to show that $X,Y$ are two random variables and $E(X\mid Y)=E(X)$, but $X$ and $Y$ are not independent.   I already know that if $X,Y$ are independent, then $E(X\mid Y)=E(X)$.
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Is there any method to show that the sum of two independent Cauchy random variables is Cauchy? I know that it can be derived using Characteristic Functions, but the point is, I have not yet learnt Characteristic Functions. I do not know anything about Complex Analysis, Residue Theorem, etc.   I would want to prove the statement only using Real Calculus. Feel free to use Double Integrals if you please.   On searching, I found . However, I was wondering if I could get some help directly on the convolution formula:  $$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx=\int_{-\infty}^\infty\frac{1}{\pi^2}.\frac{1}{1+x^2}.\frac{1}{1+(z-x)^2}dx\tag{1}$$  Here I have supposed that $X,Y$ are Independent Standard Cauchy. But I think the general formula can be derived easily after some substitutions. I need some help on how to proceed from $(1)$.  EDIT: Just as what the hint in the hyperlink said, I got the answer using that hint. However, I am not quite sure that the hint is algebraically correct. Maybe there has been some typing mistake in the book.
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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 am looking for an example of a continuous distribution function where the first moment does not exist but the characteristic function is differentiable everywhere.  Cauchy distributions do not fulfill this, as their characteristic functions are not differentiable at $0$. Does anybody here have an example?
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$$f(x,y) = \begin{cases} 2 &amp; 0&lt; x&lt; y&lt;1  \\     0 &amp;\text{otherwise} \end{cases} $$  I need to find   the density function of $y$ and conditional pdf of $X$ given $Y$.   My answer is coming out to be   $2y$ $\frac1y$
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I have a homework question that plays off my las question about the characteristic function $\chi_A(x)$.  It is:  Prove that there is a function $f$ that gives a one-to-one correspondence between $\mathcal{P}(X)$ and $2^X$.  Here the power set $\mathcal{P}(X)=\{A|A\subseteq{X}\}$ and $2^X$ is the set of all functions that map elements into the set $\{0,1\}$ Now i know that $|\mathcal{P}(X)|=2^{|X|}$, but the hint in the example mentions this:  "Now let us define a function $f$ on $\mathcal{P}(X)$ into $2^X$ by taking as the image a subset $A$ of $X$ the characteristic function of $A$," so then this mapping looks like this  $$f:\mathcal{P}(X)\rightarrow{2^X}$$ $$f(A)=\chi_A{x}$$  From here I must show injectivity and surjectivity.  So $$f(A)=f(B)\Rightarrow{A=B}$$  So by the definition of the characteristic function we can show that injectivity holds since if $$x\in{A}, \chi_A(x)=1=\chi_B(x) \text{ if } x\in{B}$$ and $$x\in{X-A}, \chi_A(x)=0=\chi_B(x) \text{ if } x\in{X-B}$$ But how do we create a surjectivity argument?
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Let $x$ and $y$ be random variables with the following joint density function: $f(x,y) = e^{-x}$ for $0&lt;x&lt; \infty$, and $0&lt;y&lt;1$  If $z= x+2y$, what is the joint density function of $x$ and $z$.  Can we first find the PDF of x and then z and then multiply the two. With that approach the answer doesn’t match
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I would like to find the characteristic function of the product of two independent brownian motions. This boils down to the characteristic function of the product of two normal random variables. This is known to follow the product-normal distribution:    but I haven't found an easy proof of this fact. This is proven making use of the Mellin transform, but this questions was in an assignment of a considerably lower level and no mention of the Mellin transform in the lectures either. I would be really happy if someone could help me.
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Maybe this is extremely simple, but i havent found a specific answer for this online.  For a sequence of independent continuous random variables $X_n$ ,$n=1,2,3,...$ , all with the same probability density function $f(x)&gt;0$ for all real $x$, and a specific real number $b$, what do the following events mean in simple terms and what is the probability of each?  a)$\lim_n\sup\{X_n \gt b\}$ b)$\lim_n\inf\{X_n \gt b\}$ c)$\lim_n\sup\{X_n \le b\}$ d)$\lim_n\inf\{X_n \le b\}$
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THE PROBLEM :     Let $X$ and $Y$ be independent and geometrically distributed with parameter $p$. Prove that $U=\min\{X,Y\}$ and $V=X-Y$ are independent. (This property characterizes geometric distributions.)   What I did was a brute-force computation of $P(U \leq u) \times P(V\leq v)$ which came out to be $$(1-q^u) \times \left(1-\frac{q^{v+1}}{1+q}\right)$$  However, computation of $P(U \leq u, V\leq v)$ is messier with lot of cases depending on the values of $u$ and $v$. I cannot get to the end. Is there any clever, nice way to solve the problem (without getting into messy calculations)? Thanks in advance.
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Supposing $X$, $Y$ and $Z$ and mutually independent real random variables, how can we prove that $X+Y$ and $Z$ are independent from the definition? If not from the definition, using $\sigma$-algebras?  I know that if $X$ and $Y$ are not independent this doesn't work so I know I have to make use of their independence in my proof but I can't figure out how. And it means that if I write the following, I'm making a mistake, and I can't find it:  $$\mathbb{P}(X+Y \leq k,Z \leq l)=\int_{\mathbb{R}} \mathbb{P}(X=x,Y \leq k-x, Z \leq l) dx=\int_{\mathbb{R}} \mathbb{P}(X=x,Y \leq k-x) \mathbb{P}(Z \leq l) dx = \mathbb{P}(Z \leq l)\int_{\mathbb{R}} \mathbb{P}(X=x,Y \leq k-x)dx= \mathbb{P}(Z \leq l) \mathbb{P}(X+Y \leq k) $$  PS: I have seen the proof here but it much more advanced stuff, kind of a sledgehammer...
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Let $X$ and $Y$ be independent random variables. Prove that $f(X)$ and $g(Y)$ are independent for any choice of measurable functions $f$ and $g$.   This sounds very obvious, but I have no idea how to approach it.  EDIT: Two random variables are independent if $\Pr\{X = x \text{ and } Y = y\} = \Pr\{X = x\} \cdot \Pr\{Y = y\}$
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What is the intuition behind the probability density function of a continuous random variable? Integrating it within two points provides the probability that is associated between two points, but if you plug a single value into $f(X=x)$, it outputs a value. What is the intuitive understanding of this value for a given x?   For example, in the above image, the value of the red curve at x=0 is 0.4. How can 0.4 be interpreted?
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The following is from the Wikipedia page on the :     By identifying a function in $2^S$ with the corresponding preimage of $1$, we see that there is a bijection between $2^S$ and $\mathcal P(S)$, where each function is the characteristic function of the subset in $\mathcal P(S)$ with which it is identified.   I don't understand the start of the sentence,"By identifying a function in $2^S$ with the corresponding preimage of $1$".  I can make out some sense out of the rest.  Please help.