We have an iid sequence with: $X_1,X_2,...$ with mean $\mu$ and variance $\sigma^2$.  We have another similar sequence: $$Y_1,Y_2,...$$.  We have a sequence $U1,U_2,...$ where, $U_n=1/\sqrt(2(n\sigma^2))(\Sigma X_j-\Sigma Y_j)$  We also have   $$W_n=1/\sqrt(n\sigma^2)(\Sigma X_j-n\mu)$$  and   $$Z_n=1/\sqrt(n\sigma^2)(\Sigma Y_j-n\mu)$$  How do we show that $\psi(U_n)(\theta)=\psi_(W_n)(\theta/\sqrt2).\psi_(Z_n)(-\theta/\sqrt2) $  ie the characteristic function of $U_n(\theta)$ evaluated at theta is equal to the product of the characteristic functions of $W_n$ and $Z_n$ evaluated at theta over root 2 and minus theta over root two repectively?  This question is about proving the characteristic function of $U_n$ can be shown as a product of the characteristic functions of $W_n$ and $Z_n$.  Approach so far:  I started trying to derive the characteristic function of Un by doing this:  $$ ψ_(U_n ) (θ)=E[e^iθX ]=∫e^iθx U_n dx =E cos⁡θX+iE sin⁡θX$$  $$=∫cos⁡θx  U_n dx+i∫sin⁡θx  U_n dx$$  $$=∫cos⁡θx  1/(√(2nσ^2 )  ) (∑X_j-∑Y_j )dx + i∫sin⁡θx 1/(√(2nσ^2 )  ) (∑X_j-∑Y_j)dx $$  But then i got stuck trying to integrate the sigmas. I also doubted that this was the correct approach given that this is a sequence of variables we are talking about.
Suppose that $U_n=\frac{1}{\sqrt{2n\sigma^2}}\left(\Sigma X_j-\Sigma Y_j\right)$,  where $X_1,X_2,\ldots$ and $Y_i,Y_2, \ldots$ are i.i.d. sequences of random variables with mean $\mu$ and variance $\sigma^2$ and the sequences $\{X_i\}$ and $\{Y_i\}$ are independent.  Then, for all real numbers $x$, show that, as $n\to \infty$,  $$\mathbb{P}(U_n\leq x) \to  \int_{-\infty}^{x} \frac{1}{(\sqrt(2\pi))} \:e^{-\frac 12{u^2}}\mathrm du $$ Approach  I tried to carry out the integration on the right, but this does not integrate nicely. It ends up involving an error function, so I figured this was the wrong was to go.  Should I be looking at convergence in probability to answer this question? Is Chebyshev's inequality relevant here? If so, how and how does that relate to the integral?
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Error term in equally spaced nodes Prove that $\mathcal{P}(A)⊆ \mathcal{P}(B)$ if and only if $A⊆B$. Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    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