\section{\texorpdfstring{Repetition, NS hierarchy}{Repetition, NS hierarchy}}
\vspace{5mm}
\large

\subsection{Abbreviations}
\begin{itemize}
	\item TM - turing machine.
	\item DTM - deterministic turing machine.
	\item NTM - non-deterministic turing machine.
	\item DS - DSPACE
	\item NS - NSPACE
	\item PS - deterministic polynomial space
	\item DT - DTIME
	\item NT - NTIME
	\item QBF - quantifiable boolean formula
	\item CNF - conjunctive normal form
	\item PH - polynomial hierarchy
	\item PTM - Probabilistic TM
	\item r.v. - Random variable
	\item RP - Randomized Polynomial time
	\item ZPP - Zero-error Probabilistic Polynomial time
\end{itemize}

\begin{theorem}[NT and DS relation]\label{nt_ds_rel}
	\[ \forall f:\N \to \N: NT(f(n)) \subseteq DS(f(n)) \]
\end{theorem}

\begin{proof}
Construct TM with following algorithm:
\begin{enumerate}
	\item k = 1
	\item do
	\item foreach $y$ branch $|y| = k$
	\item Simulate M(x) by $y$
	\item If(Acc) accept
	\item k++;
	\item until all simulations of M(x) by $y$ that rejected
		// if all branches rejected $\Rightarrow$ there is no accepting computation
	\item reject
\end{enumerate}

M works in time $g(n) = \bigO(f(n))$, computation in step 4. will do at most g(n) steps.

Space:\\
1) to store $k$ we need $g(n)$ of space, $ k \leq g(n) $. \\
2) To simulate $M(x)$ by $y$ we need $\bigO(g(n))$ space.

Correctness: \\
We want constructed DTM to accept the same language as initial NTM.
If there is an accepting branch of NTM, DTM also accepts by some branch $y$.
Otherwise we reach $k = g(n)$ and DTM rejects.

TM cannot loop forever by the definition of time complexity (we restrict the languages to recursive).

Technical details:\\
Constructed TM should have 2 tapes:\\
a) working tape\\
b) tape to store $y$ (vector that encode branch of NTM)
\end{proof}

\begin{note}
	If we are constrained to use single tape, any $k$ tape machine can be compressed to 1 tape machine. We need 2 steps:

\begin{enumerate}
	\item compress $\Sigma \to \Sigma^k$
	\item How to deal with many heads of the original TM? $\Rightarrow$ simulate 1 step of the original machine by many steps in new TM.
		In total, time complexity is $\bigO(n^2)$ and space complexity is $\bigO(n)$.
\end{enumerate}
\end{note}

\begin{theorem}[NS and DT relation]\label{ns_dt}
	\[\forall f:\N \to \N, \forall L \in NS(f(n)) \Rightarrow \exists c_L: L \in DT\left(c_L^{f(n)}\right) \]
\end{theorem}
\begin{proof}
	All accepting configuration leaves are bounded by the \# of configurations. However, \# of all paths $2^{c_L^{f(n)}}$.
	How to avoid double exponent? $\Rightarrow$ Perform BFS in configuration graph.
	\# of edges could be quadratic , however
	\[ \left(c_L^{f(n)}\right)^2 = \left(c_L^2\right)^{f(n)} \]
	$c_L^2$ is another constant.
\end{proof}

\begin{definition}[Universal TM]
Input: $(x,y)$, where $x$ is Godel number of TM to be simulated. $y$ is input of TM.\\
Alphabet: $\Sigma = \{ 0, 1 \} $.\\
Working tapes: 3
\begin{enumerate}
	\item Tape with transition table of $M_x$.
	\item Tape with current state $q$
	\item Working tape
\end{enumerate}
If $S(n)$ is space used by $M_x \Rightarrow$ we need $\max(\lceil \log(t) \rceil, S(n), |x|)$.\\
TODO what is $t$?
\end{definition}

\begin{observation}
	If initial TM has $k$ tapes and time complexity is $T(n)$ we can compress to 2 tapes with a cost of time complexity being $\bigO(T(n) \log(n))$. \\
	// TODO write proof (moving blocks on tape)

	Therefore simulating TM with multiple tapes could be reduces to 2 tapes, then simulate on Universal TM.
	Time complexity of Universal TM is dominated by finding the transition $\Rightarrow \bigO(|x| \cdot T(n))$.
\end{observation}

\begin{definition}[Space constructible function]
	Function $f$ is \emph{Space constructible} $\iff \exists$ TM with unary alphabet which marks exactly $|f(n)|$ cells on the working tape. E.g. $\log, e^x,$ polynomials.
\end{definition}

\begin{theorem}[Space hierarchy]\label{s_hier}
	Let $S_1, S_2$ are space constructible functions and
	\[ S_1 \in o(S_2) \Rightarrow DS(S_1(n)) \subsetneq DS(S_2(n)) \]
\end{theorem}
\begin{proof}
	Construction by Cantor diagonalization method $\Rightarrow$ find a language that is different from all in $\bigO(S_1)$.
\end{proof}

\begin{definition}[Time constructible function]
	Function $f$ is \emph{Time constructible} $\iff \exists$ TM that does exactly $|f(n)|$ steps. You can think of it as alarm clock.
\end{definition}

\begin{observation}
	Every time constructible function is space constructible.
\end{observation}

\begin{theorem}[Time hierarchy]\label{t_hier}
	Let $S_1, S_2$ are time constructible functions and
	\[ T_1 \cdot \log(T_1(n)) \in o(T_2) \Rightarrow DT(T_1(n)) \subsetneq DT(T_2(n)) \]

	Note that $\log(T_1(n))$ is required because of $k$ to 2 tapes compression.
\end{theorem}
\begin{proof}
	Construction by Cantor diagonalization method $\Rightarrow$ find a language that is different from all in $\bigO(T_1)$.
\end{proof}

\begin{theorem}[Savic]\label{savic}
	Under some mild assumptions $($space constructibility, functions bigger than $\log(n))$ following statement is true:
	\[ NS(f(n)) \subseteq DS(f^2(n)) \]

\end{theorem}
\begin{proof}
	Find a path in configuration path. We cannot use neither BFS not DFS as the complexity is linear in edges which is exponential comparing to input.
	Therefore we use recursive algorithm which for all states $K$ reachable by a path of length
	\[ \frac{c_L^{f(n)}}{2^i}\]
	tries to find a path from $C_{init} \to K \to C_{accept}$.

	As we divide path by 2 at every recursive call, recursion tree height is equal to $\log_2(n)$.
	Therefore time complexity is
	\[ \log_2(c_L^{f(n)}) = f(n) \cdot \log_2(c_L) \]

	On each level of recursion we have to store $C_{init}, K, C_{accept}$ which requires $\bigO(f(n))$ space.
	Having $\bigO(f(n))$ levels, total space complexity is $\bigO(f^2(n))$.
\end{proof}

\begin{note}
	Time version of Savic would imply $P = NP$.
\end{note}

\subsection{NS hierarchy}

% todo Cepek has kinda different proof
\begin{lemma}[Translation lemma]\label{transl_space}
	Let $S_1(n), S_2(n), f(n)$ be space constructible functions, also
	\[ S_2(n) \geq n, f(n) \geq n \]
	Then
	\[ NS(S_1(n)) \subseteq NS(S_2(n)) \Rightarrow NS(S_1(f(n))) \subseteq NS(S_2(f(n))) \]

	Lemma allows to replace $n \to f(n)$.

	We can also prove Translation Lemma for $DS, NT, DT$.
\end{lemma}
\begin{proof}
	Let $L_1 \in NS(S_1(f(n)))$ arbitrary, $L_1$ is recognized by NTS $M_1$ in space $S_1(f(n))$.
	We want to prove that $L_1 \in NS(S_2(f(n)))$ by constructing $L_2 \in NS(S_1(n))$ using padding.

	Define $L_2 := \{ x\triangle^i | M_1 $ accepts $x$ in space $S_1(|x| + i) \} $.
	Where $\triangle$ is a new symbol, that $\notin $ initial $\Sigma$.

	Algorithm of $M_2, L(M_2) = L_2$ on input $x\triangle^i$ is the following:
\begin{enumerate}
	\item Mark $S_1(|x| + i)$ cells on tape.
	\item Simulate $M_1$.
	\item if($M_1(x)$ accept $\land$ did not use more space than marked in 1) then accept
\end{enumerate}
	From the construction, $M_2$ recognizes $L_2$ using $S_1(n)$ space.
	Consequently, $L_2 \in NS(S_1(n))$.
	Combining with assumption of the lemma
	\[ L_2 \in NS(S_2(n)) \]
	And there exists a TM $M_3$ that recognizes $L_2$ using $S_2(n)$ space.

	The last step is to construct $M_4$ that recognizes $L_1$ using $S_2(f(n))$ space.
	$M_4$ has 2 working tapes and has following algorithm ($M_4$ on input $x$):
\begin{enumerate}
	\item Mark $f(n)$ cells on 1st tape.
	\item Mark $S_2(f(n))$ cells on 2nd tape. Using 1st tape as "input".
	\item foreach $x\triangle^i, i = 0, 1, \ldots $ \\
		simulate $M_3$ on input $x\triangle^i$.
		If head of $M_3$ is inside $x$ $M_4$s head is in the same place.
		Otherwise, head of $M_3$ is inside padding, so $M_4$ uses counters to track $M_3$s position.
		Counter has length at most $1 + \log i$.
	\item if($M_3(x)$ accept) then $M_4$ accept.
	\item else if($1 + \log(i) \leq S_2(f(n))$) then $++i$; \\
		Counter did not overflow and is less than $S_2(f(n))$.
	\item else if($1 + \log(i) > S_2(f(n))$) then reject;
\end{enumerate}

If $M_4$ accepted input $x$, $M_3$ accepted $x\triangle^i$ for some $i \Rightarrow x\triangle^i \in L_2 \Rightarrow M_1$ accepts $x \Rightarrow x \in L_1$.

On the other hand, if $x \in L_1 \Rightarrow x\triangle^i \in L_2$ for $i = f(|x|) - |x|$ therefore counter $i$ requires
\[ \log(f(|x|) - |x|) \leq S_2(f(|x|)) \]
space.

\end{proof}

\begin{theorem}[NS hierarchy for polynomials]
	Let $\varepsilon > 0,\, r > 1$. Then
	\[ NS(n^r) \subsetneq NS(n^{r + \varepsilon}) \]
\end{theorem}
\begin{proof}
	From the density of rationals:
	\[ \exists s,t \in \N: r \leq \frac{s}{t} \leq \frac{s + 1}{t} \leq r + \varepsilon \]
	It is sufficient to prove that:
	\[ NS(n^{\frac{s}{t}}) \subsetneq NS(n^{\frac{s + 1}{t}}) \]

	Assume by contradiction
	\[ NS(n^{\frac{s+1}{t}}) \subseteq NS(n^{\frac{s}{t}}) \]
	Now we use \cref{transl_space} for
	\[ S_1 = n^{\frac{s+1}{t}}, S_2 = n^{\frac{s}{t}}, f(n) = n^{(s + i)t}, i = 0, 1 \ldots, t \]
\begin{note}
\begin{note}
\end{note}

\end{note}

	We get
	\[ \forall i: NS((n^{(s + i)t})^{(s + 1)/t}) \subseteq NS((n^{(s + i)t})^{s/t}) \Rightarrow \forall i: NS(n^{(s + i)(s + 1)}) \subseteq NS(n^{(s + i)s}) \]
	Now we write for all $i$:
	\begin{itemize}
		\item $i = 0:\, NS(n^{s(s + 1)}) \subseteq NS(n^{s^2}) $
		\item $i = 1:\, NS(n^{(s + 1)(s + 1)}) \subseteq NS(n^{(s + 1)s}) $
		\item \ldots
		\item $i = s:\, NS(n^{2s(s + 1)}) \subseteq NS(n^{(2s)s}) $
	\end{itemize}
	From the exponents we conclude that for every inequality right side is a subset of left side.
	So we get a chain of subsets and can use Savic theorem to get a contradiction:
	\[ NS(n^{(s + 1)2s}) \subseteq NS(n^{s^2}) \stackrel{Savic}{\subseteq} DS(n^{2s \cdot s}) \stackrel{space H}{\subsetneq} DS(n^{2s \cdot s + 2s}) \subseteq NS(n^{2s(s + 1)}) \]
	As the beginning and the end are the same sets and the chain of subsets has strict inclusion.

\end{proof}


\section{Fission probability calculation.}

\hspace{1.0em}The fission decay channel (only for nuclei with $A > 65$)
is taken into account as a competitor for fragment and photon evaporation
channels.

\subsection{The fission total probability.}
hspace{1.0em}The fission probability (per uni time) $W_{fis}$
in the Bohr
and Wheeler theory of fission \cite{BW39} is proportional to the level
density $\rho_{fis}(T)$ ( approximation Eq. ($\ref{SFE13}$) is used) at
the saddle point, i.e.
\begin{equation}
\begin{array}{c}
\label{FP1}W_{fis}=\frac{1}{2\pi \hbar \rho_c(U_c)}\int_{0}^{U_f-B_{fis}}
\rho_{fis}(U_f-B_{fis}-T)dT =\\
=\frac{1 + (C_f - 1)\exp{(C_f)}}{4\pi a_{fis} \exp{(2\sqrt{a_cU_c})}},
\end{array}
\end{equation}
where $U_f= E^{*} - \Delta_f$ and pairing energy
\begin{equation}
\label{FP2} \Delta_{f} = \kappa \frac{14}{\sqrt{A}} \ [MeV]
\end{equation}
In Eq. ($\ref{FP1}$)  $B_{fis}$ is the fission barrier height.
The value of $C_f = 2\sqrt{a_{fis}(U_f - B_{fis})}$ and $a_c$, $a_{fis}$ are
the level density parameters of the compound and of the fission saddle point
nuclei, respectively.

The value of the level density parameter is large at the saddle point,
when excitation energy is given by initial excitation energy minus the
fission barrier height, than in the ground state, i. e. $a_{fis} > a$.
$a_{fis} = 1.08 a$ for $Z < 85$, $a_{fis} = 1.04 a$ for $Z \geq 89$ and
$a_f=a[1.04+0.01(89.-Z)]$ for $85 \leq Z < 89$ is used.

\subsection{The fission barrier.}

\hspace{1.0em}
The fission barriers are determined as differences
between the saddle-point and
 ground state masses. In the general case fission barriers are functions
 of the charge $Z$, atomic mass number $A$, excitation energy of fissioning
 nuclei $E^{*}$ and their angular momenta $L$ and their deformations
 $\alpha$.

 Shell structure effects play a role at the fission
barrier. The height of fission barrier can be aprroximated as
\begin{equation}
\label{FP3}B_{fis} = B^{0}_{fis} + \Delta_{Shell} + \Delta_{SP},
\end{equation}
where $B^{0}_{fis}$ is the so-called liquid drop component of the
fission barrier, $\Delta_{Shell}$ is the shell correction to the mass of a
nucleus in the ground state and $\Delta_{SP}$ is the shell correction to
the mass of nucleus in the saddle point.  The last correction is very
important for actinide nuclei. It leads to a double-humped shape of the
fission barrier.

There are many models for fission barriers: the phenomenological approach
of Barashenkov et al. \cite{Barash73}, the semiphenomenological
approach of Barashenkov and Gereghi \cite{Barash77}, the liquid-drop model (LDM)
with Myers and Swiatecki's parameters \cite{MS67}, the LDM with Pauli and Ledergerber's
 parameters \cite{PG71}, the single-Yukawa modified LDM of Krappe and Nix
 \cite{KN73}, the Yukawa-plus-exponential modified LDM \cite{KNS79}, the subroutine
 BARFIT of Sierk \cite{Sierk86} which provides macroscopic fission barriers of
 rotating nuclei in the Yukawa-plus-exponential modified LDM \cite{KNS79},
 double-humped fission barriers for transuranium nuclides as proposed in
 \cite{KIF80}, giving a fixed
 input value for single-humped fission barriers $B_{fis}$
 and giving fixed input values $B^A_{fis}$ and $B^B_{fis}$ for
 double-humped fission barriers.

We use simple semiphenomenological
approach was suggested by Barashenkov and Gereghi \cite{Barash77}. In their
approach fission barriers $B_{fis}(A,Z)$ are approximated by
\begin{equation}
\label{FP4} B_{fis} = B^{0}_{fis} + \Delta^{C}_{Shell}  + \Delta^{C}_{Pair} +
\delta(A,Z),
\end{equation}
where shell and pairing corrections  for Cameron's
liquid drop mass formula \cite{CAM57}. $\delta(A,Z) = 0$ for even $Z$
and even $N = A - Z$, $\delta(A,Z) = 1.248$ \ MeV for odd $A$ and
$\delta(A,Z) = 2.496$ \ MeV for odd $Z$ and odd $N$ were suggested.

According to this prescription fission barrier heights $B^{0}_{fis}(x)$
vary with the fissility parameter $x$. $B^{0}_{fis}(x)$ is
given by
\begin{equation}
\label{FP5} B^{0}_{fis}(x) = a_{S}A^{2/3} 0.83(1 - x)^3
\end{equation}
for $2/3 \leq x \leq 1$ and
\begin{equation}
\label{FP6} B^{0}_{fis}(x) =a_{S}A^{2/3} 0.38(3/4 -x)
\end{equation}
for $1/3 \leq x \leq 2/3$.
The fissility parameter $x$ is given by
\begin{equation}
\label{FP7} x = \frac{E^0_C}{2E^0_S}= \frac{(a_C/2a_S)Z^2/A}
{\{1 - k[(N-Z)/A]^2\}},
\end{equation}
where $E^{0}_C$ and $E^{0}_S$ are the Coulomb and surface energies of
a spherical nucleus, respectively.

The liquid drop model parameters $a_S = 17.9439$\ MeV, $a_C=0.7053$ \ MeV
and $k = 1.7826$ taken from \cite{CS63} paper.

The fission barrier heights are functions of the excitation energy. We use the
empirical relation proposed by \cite{Barash73} to estimate the dependence of
$B_{fis}$ on $E^{*}$:
\begin{equation}
\label{FP8} B_{fis}(E^{*}) =\frac{B_{fis}}{1+\sqrt{(\frac{E^{*}}{2A})}}.
\end{equation}


\documentclass[11pt,a4paper]{article}
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\title{Lecture 30: Lift due to Potential Flows in 2D}
\date{\displaydate{date}}
\newdate{date}{01}{11}{2018}
\author{}

\begin{document}
\maketitle

\section*{overview}
Lift on an aerofoil occurs due to a net circulation associated with the aerofoil. Potential flow theory can be used to determine the flow field and hence the lift associated with the aerofoil under a given uniform flow.

\section{Flow around a circular cylinder with circulation}

The complex potential for flow around a translating circular cylinder (with velocity $U$) with a given circulation ($\kappa$) is
\begin{align*}
&F(z) = \describe{translation}{U\bigg( z + \frac{a^2}{z}\bigg)} - \describe{circulation}{\frac{i\kappa}{2\pi}\log z}
\end{align*}
Let the real part of this complex potential be the velocity potential $\Phi$
\begin{align*}
&\Phi  = Ur\cos\phi\bigg(1+\frac{a^2}{r^2} \bigg) + \frac{\kappa \phi}{2\pi}\\
\RA& u_\phi\Big|_{r=a} = \frac{\kappa}{2\pi a} - U \sin\phi
\end{align*}
This is the velocity field in the frame of reference of the moving cylinder. Now we want to find out the stagnation points of this flow. For $\kappa = 0$, they lie at $\phi = 0, \pi$ (as can be seen from the expression for $u_\phi$). In the presence of circulation, however, the stagnation points move closer and eventually merge and move into the fluid giving rise to homoclinic-type streamlines. It must be noted that vortex opposes the uniform flow on one side of the cylinder, and strengthens it on the other side. This difference in velocities leads to a pressure gradient in accordance with Bernoulli's theorem. This is the lift force.

\section{Flow around a translating elliptic cylinder with circulation}

The complex potential for this case is given by
\begin{align*}
&W(\zeta) = U\bigg(\zeta e^{-i\alpha} + \frac{(a+b)^2}{4\zeta} e^{i\alpha}\bigg) - \frac{i\kappa}{2\pi} \log\zeta
\end{align*}
This is the potential for a circular cylinder. Using the conformal mapping
\begin{align*}
&\zeta = \frac{1}{2}(z+\sqrt{z^2-c^2}), \tag{$c^2 = a^2 - b^2$}
\end{align*}
we map it to an ellipse. This gives us the complex potential in the $t (=t_1+it_2)$ plane
\begin{align*}
&\tilde{w}(t) =\\
& \frac{U}{2}\bigg(c (\cosh t + \sinh t) e^{-i\alpha} + (a+b)^2\frac{e^{i\alpha}}{c} (\cosh t - \sinh t)\bigg) - \frac{i\kappa}{2\pi}\log(\cosh t + \sinh t)
\end{align*}
From this potential, we can evaluate the tangential velocity at the surface of the ellipse given by $\frac{d\tilde{w}(t)}{dt}$ @ $t=t_{1_0} + it_2$
\begin{align*}
&\frac{d\tilde{w}(t)}{dt} = iUc\bigg(e^{t_{1_0}}\sin(t_2-\alpha) - \frac{\kappa}{2\pi c U} \bigg)
\end{align*}
The behaivour of stagnation points and corresponding change in the lift force as the circulation is increased from zero, is similar to the case of the sphere. When circulation is zero, the stagnation points are located at $\pi \pm \alpha$. As circulation is increased, the points move closer to each other. Consider now, a thin ellipse $t_{1_0} = 0$. Potential flow rounds the edges of such a shape with infinite velocity. As circulation is increased, one of the stagnation point eventually reaches the edge at $t_2 = 0$. Thus velocity at the trailing edge becomes finite and we get a net lift force due to the circulation associated with the thin cylinder.

\section{Appendix}
\subsection{Potential flow in multiply connected domain}

\end{document}


\documentclass{beamer}

\mode<presentation>
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  \usetheme{default}      % or try Darmstadt, Madrid, Warsaw, ...
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\usepackage[english]{babel}
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\setlength\tabcolsep{1.5pt}

\begin{document}
\frame{
\frametitle{Contents}
\tableofcontents
}

\section{Estimation from a damage function}
\frame{
\frametitle{Introductory thoughts}

\begin{itemize}
\item The impact of CO$_2$ on impacts has a much shorter half-life than of CO$_2$ in the atmosphere because of adaptation-- but we know exactly how long that timescale is: a Bartlett kernel-weighted 30 years.
\item The impact of a step change in temperature produces a short-term effect, in the immediate year, and a long-term effect, 30 years later.
\item The damage function we estimate for IAMs should incorporate this transition, both in its estimation and in the information we provide to IAMs.
\end{itemize}
}

\frame{
\frametitle{A model of impacts}

Changes in CO$_2$ leads to changes in temperature according to a scientifically assumed transfer function, $p_t = (1 - e^{-t / 2.8}) e^{-t / 400}$: $T_t = C_t \ast p_t$.  (Throughout, $\ast$ is the convolution operator.)

Assume that impacts are generated by
\[
y_t = \left[f(T_t) - a (f(T_t) \ast b_t)\right] (\text{GDPpc}_t)^\gamma
\]
\begin{itemize}
\item $f(T_t)$ is an instantaneous damage function, generally of the form $\beta_1 T_t + \beta_2 T_t^2$.
\item $b_t$ is the Bartlett kernel, and $a$ is the degree of temperature-driven adaptation.
\item The last term provides a measure of elasticity of damages with income.
\end{itemize}
}

\frame{
\frametitle{Estimated damage function}

\includegraphics[width=\textwidth]{damagefunc2.pdf}
}

\frame{
\frametitle{Bayesian fitted parameters}

\includegraphics[width=\textwidth]{bayes.pdf}
}

\frame{
\frametitle{Results}

\begin{tiny}
\begin{tabular}{lllrrrrrrrr}
RCP & SSP & Monetization & 0 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 \\
4.5 & SSP4 & VSL ag02 popavg & 49.765215 & 20.090173 & 11.188628 & 7.319126 & 5.2335889 & 3.9565314 & 3.1067942 & 2.5073709 \\
4.5 & SSP4 & VSLY popavg & 27.076281 & 11.585834 & 6.923858 & 4.851441 & 3.7002694 & 2.9724734 & 2.4727046 & 2.1093082 \\
4.5 & SSP4 & VSLY scaled & 4.846833 & 2.353378 & 1.527805 & 1.128928 & 0.8938308 & 0.7389402 & 0.6293463 & 0.5477955 \\
4.5 & SSP4 & VSL ag02 scaled & 10.848195 & 5.175614 & 3.320996 & 2.429363 & 1.9066906 & 1.5647519 & 1.3247003 & 1.1474737 \\
4.5 & SSP5 & VSL ag02 popavg & 22.639750 & 12.790459 & 9.110520 & 7.113661 & 5.8206464 & 4.9042500 & 4.2186996 & 3.6866908 \\
4.5 & SSP5 & VSLY popavg & 11.801313 & 7.208263 & 5.423449 & 4.413440 & 3.7358821 & 3.2413275 & 2.8619107 & 2.5608733 \\
4.5 & SSP5 & VSLY scaled & 3.848509 & 2.133306 & 1.521382 & 1.197915 & 0.9917263 & 0.8470242 & 0.7393777 & 0.6560325 \\
4.5 & SSP5 & VSL ag02 scaled & 11.471691 & 5.608791 & 3.721099 & 2.799517 & 2.2450104 & 1.8726067 & 1.6050818 & 1.4037624 \\
8.5 & SSP4 & VSL ag02 popavg & 112.962731 & 41.333858 & 21.802857 & 13.955777 & 9.9695436 & 7.6319429 & 6.1235417 & 5.0814877 \\
8.5 & SSP4 & VSLY popavg & 54.144549 & 20.820391 & 11.622419 & 7.842783 & 5.8696309 & 4.6790999 & 3.8891723 & 3.3288767 \\
8.5 & SSP4 & VSLY scaled & 10.365246 & 4.535453 & 2.715188 & 1.892462 & 1.4381250 & 1.1560006 & 0.9663930 & 0.8312908 \\
8.5 & SSP4 & VSL ag02 scaled & 22.627928 & 10.012589 & 5.961467 & 4.101235 & 3.0711859 & 2.4350453 & 2.0115973 & 1.7132506 \\
8.5 & SSP5 & VSL ag02 popavg & 43.699053 & 21.802650 & 14.734155 & 11.293759 & 9.2161600 & 7.8077671 & 6.7840025 & 6.0041636 \\
8.5 & SSP5 & VSLY popavg & 20.489387 & 11.096381 & 7.938798 & 6.331466 & 5.3222833 & 4.6156765 & 4.0879440 & 3.6765711 \\
8.5 & SSP5 & VSLY scaled & 6.789311 & 3.387298 & 2.244069 & 1.682805 & 1.3488154 & 1.1277918 & 0.9711668 & 0.8546137 \\
8.5 & SSP5 & VSL ag02 scaled & 19.423931 & 8.974785 & 5.584819 & 3.984682 & 3.0697847 & 2.4871389 & 2.0886172 & 1.8013216 \\
\end{tabular}
\end{tiny}
}

\frame{
\frametitle{Calculating an SCC (one slide for MG)}

\begin{enumerate}
\item We estimate a global damage function: total monetized costs as they vary with temperature change from the baseline.
\begin{center}\includegraphics[width=.6\textwidth]{damagefunc2.pdf}\end{center}
\item The damage function includes the effect of temperature adaptation (minor here) and income adaptation (huge).
\item We calculate $(A)$ costs for each year according to an RCP scenario, and $(B)$ costs for a 1 tonne boost in 2017 above that scenario.
\item The SCC is the present discounted value of $(B) - (A)$.
\end{enumerate}
}

\section{Reduced-form estimation}
\frame{
\frametitle{Calculating an SCC}
Let $x_t$ be a stream of CO$_2$ emissions and $y_t$ be the stream of impacts.

Let $f(t)$ be an impulse response function which describes how a single GT jump in CO$_2$ produces a stream of impacts.  Then, $y_t = x_t \ast f(T)$, the result of a convolution.

\begin{enumerate}
\item We assume that each unit increase in CO2 has an impact that starts near 0, rises rapidly, and slowly decays.  It also varies with average global temperature $T$.
\item The entire stream of impacts from CO2 is the just the sum of scaled and translated copies of this impulse response.
\item We calculate the coefficients that define that impulse response.
\item We calculate the NPV of the impulse response.
\end{enumerate}
}

\frame{
\frametitle{A proposed structural form}

Suppose that $f(T) = \sum_{k=0}^K \left(\beta_{k0} + \beta_{k1} T\right) t^k e^{-t / \tau}$, where $\tau$ is the residence time of CO$_2$ in the atmosphere, 400 yr under IPCC or 77 yr under DICE.

Then,
\begin{align*}
y_t &= \alpha + \sum_{s=0}^\infty x_{t - s} \sum_{k=0}^K \left(\beta_{k0} + \beta_{k1} T_{t-s}\right) s^k e^{-s / \tau} + \epsilon_t \\
&= \alpha + \sum_{k=0}^K \beta_{k0} \sum_{s=0}^\infty x_{t - s} s^k e^{-s / \tau} + \sum_{k=0}^K \beta_{k1} \sum_{s=0}^\infty x_{t - s} T_{t-s} s^k e^{-s / \tau} + \epsilon_t
\end{align*}

This is just a weighted sum of recent CO$_2$ emissions as the predictors for a regression.
}

\frame{
\frametitle{Calculating an SCC}
We are interested in the discounted sum of impacts.  This can be calculated as,

\begin{align*}
SCC_t &= \sum_{u = 0}^\infty e^{-\delta u} y_{u}(\delta_u) \\
&= \sum_{u = 0}^\infty e^{-\delta u} \left[\sum_{k=0}^K \beta_{k0} \sum_{s=0}^\infty \mathbf{1}\{u-s = 0\} s^k e^{-s / \tau} +\right. \\
&\hspace{6em} \left.\sum_{k=0}^K \beta_{k1} \sum_{s=0}^\infty \mathbf{1}\{u-s = 0\} T_{u-s} s^k e^{-s / \tau}\right] \\
&= \sum_{u = 0}^\infty e^{-\delta u} \left[\sum_{k=0}^K \beta_{k0} u^k e^{-u / \tau} + \sum_{k=0}^K \beta_{k1} T_t u^k e^{-u / \tau}\right] \\
\end{align*}
}

\end{document}