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function to fit to a nonlinear function. The Fit \ function is similar, but it will fit to a linear combination of functions.\ \>", "Text", CellChangeTimes->{{3.4336439943444233`*^9, 3.433644019327944*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"FindFit", "[", RowBox[{"data", ",", RowBox[{"a", " ", RowBox[{"Exp", "[", RowBox[{ RowBox[{"-", "a"}], " ", "x"}], "]"}]}], ",", "a", ",", "x"}], "]"}], ";"}]], "Input", CellChangeTimes->{{3.433644033800477*^9, 3.43364410578375*^9}, 3.434238661536029*^9}], Cell[BoxData[ RowBox[{ RowBox[{"FindFit", "::", "\<\"fdssnv\"\>"}], RowBox[{ ":", " "}], "\<\"Search specification \\!\\(2\\) without variables should \ be a list with 1-4 elements. \\!\\(\\*ButtonBox[\\\"\[RightSkeleton]\\\", \ ButtonStyle->\\\"Link\\\", ButtonFrame->None, \ ButtonData:>\\\"paclet:ref/FindFit\\\", ButtonNote -> \ \\\"FindFit::fdssnv\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{{3.434238655049803*^9, 3.434238671954298*^9}}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[Tau]", "=", FractionBox["1", ".0688807"]}]], "Input", CellChangeTimes->{{3.43364468582028*^9, 3.433644720679986*^9}}], Cell[BoxData["14.51785478370574`"], "Output", CellChangeTimes->{{3.4336446937399883`*^9, 3.4336447211309843`*^9}, 3.433967131374997*^9, 3.433971395495776*^9, 3.434024818386736*^9, 3.434238700331607*^9}] }, Open ]], Cell["\<\ This isn't bad for an estimate. We expect something like 20 ( = 2\[Mu]s), so \ we're at least in the right ball park.\ \>", "Text", CellChangeTimes->{{3.433966469283228*^9, 3.433966489970251*^9}}] }, Closed]], Cell[CellGroupData[{ Cell["Next-to-Preliminary Data Analysis", "Section", CellChangeTimes->{{3.433966446272084*^9, 3.433966447561256*^9}, { 3.433971984142479*^9, 3.433971988205412*^9}}], Cell[TextData[{ "Here we'll do some of our heavy lifting. Hopefully it works out properly. \n\ We would like to do a ", StyleBox["maximum likelihood", FontWeight->"Bold"], " fit to the form of two exponentials plus a constant background. Note: be \ careful with normalization!" }], "Text", CellChangeTimes->{{3.433966457812175*^9, 3.4339664655777273`*^9}, { 3.4339664970110703`*^9, 3.433966568937687*^9}}], Cell[CellGroupData[{ Cell["Constants", "Subsection", CellChangeTimes->{{3.433966554303071*^9, 3.4339665551036177`*^9}}], Cell["\<\ We shall take the ratio of positive to negative muons to be a constant.\ \>", "Text", CellChangeTimes->{{3.433966558663732*^9, 3.433966585303824*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Estimate of background", "Subsection", CellChangeTimes->{{3.433966451171392*^9, 3.433966455298665*^9}}], Cell["\<\ We can estimate the background using the singles counts. We assume the \ singles counts is much greater than the subset of muon candidates (doubles). \ We can read off the total number of singles from summing our counts and we \ can read off the total time in the dataout file (it's the last number of the \ file). Note that the dataout files also have a \"total count\" before this \ number, but htis seems to be off by roughly a factor of two. The singles count has a large error, maybe on the order of (plus or) minus \ 1000, but this is fine for estimating the rate.\ \>", "Text", CellChangeTimes->{{3.433966960102694*^9, 3.433967107588887*^9}, { 3.433967459797607*^9, 3.433967494690008*^9}, {3.4339675502064*^9, 3.433967565186102*^9}}], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{ RowBox[{"time", "=", "523089"}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"singlescounts", " ", "=", " ", "99855600"}], ";"}], "\[IndentingNewLine]", RowBox[{"rate", " ", "=", " ", RowBox[{"N", "[", RowBox[{"singlescounts", "/", "time"}], "]"}]}]}], "Input", CellChangeTimes->{{3.4339674148231153`*^9, 3.433967432843793*^9}, { 3.43396750870543*^9, 3.433967529188383*^9}}], Cell[BoxData["190.8960043128416`"], "Output", CellChangeTimes->{3.4339675295234528`*^9, 3.433971397846283*^9, 3.434024820417342*^9, 3.434238704578376*^9, 3.434240367639482*^9, 3.4342405308288803`*^9, 3.434241572321515*^9, 3.434367752768813*^9}] }, Open ]], Cell["\<\ I'll estimate the background using the Poisson distribution. The probability \ for a double event, given a single event, is\ \>", "Text", CellChangeTimes->{{3.433970270624486*^9, 3.4339702720164623`*^9}, { 3.433970508742114*^9, 3.43397051983997*^9}, {3.4339710311339483`*^9, 3.433971065616871*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"Prob", RowBox[{"(", RowBox[{"double", "|", "single"}], ")"}]}], " ", "=", " ", RowBox[{"PDF", "[", RowBox[{ RowBox[{"PoissonDistribution", "[", RowBox[{"time", "*", "rate"}], "]"}], ",", "2"}], "]"}]}]], "DisplayFormula", CellChangeTimes->{{3.433971175528286*^9, 3.433971219899034*^9}}], Cell["Then the total number of expected doubles is", "Text", CellChangeTimes->{{3.4339712241601048`*^9, 3.433971235721401*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"N", RowBox[{"(", RowBox[{"bg", " ", "doubles"}], ")"}]}], " ", "=", " ", RowBox[{"N", RowBox[{"(", RowBox[{"bg", " ", "singles"}], ")"}], "\[Times]", "P", RowBox[{"(", RowBox[{"double", "|", "single"}], ")"}]}]}]], "DisplayFormula", CellChangeTimes->{{3.433971242067504*^9, 3.433971268122892*^9}}], Cell["\<\ I feel like this is a really rough estimate, but it should be good enough as \ an rough idea. Don't forget to divide by 250 bins.\ \>", "Text", CellChangeTimes->{{3.433971269730715*^9, 3.4339712880974207`*^9}, { 3.433971439116589*^9, 3.4339714441701107`*^9}}] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Unbinned fit to a single exp", "Section", CellChangeTimes->{{3.434033365096834*^9, 3.434033368516965*^9}, { 3.434145476603672*^9, 3.43414547963483*^9}, {3.4342416308111477`*^9, 3.434241634569881*^9}, {3.434311202713077*^9, 3.434311204571538*^9}}], Cell[CellGroupData[{ Cell["Faux Signal", "Subsection", CellChangeTimes->{{3.434241618781361*^9, 3.434241619910142*^9}}], Cell[TextData[{ "Descriptions:\n* ", StyleBox["listlength", FontWeight->"Bold"], " is the length of the list\n* ", StyleBox["onetolength", FontWeight->"Bold"], " is a list of numbers from 1 to [", StyleBox["listlength", FontWeight->"Bold"], "]\n* ", StyleBox["thelifetime", FontWeight->"Bold"], " is the lifetime, i.e. 1/decay rate in the exponential\n* ", StyleBox["dexp", FontWeight->"Bold"], " is the decaying exponential function, properly normalized\n* ", StyleBox["perfectdata", FontWeight->"Bold"], " is a list of [", StyleBox["listlength", FontWeight->"Bold"], "] perfect datapoints\n* ", StyleBox["mydata", FontWeight->"Bold"], " is the ", StyleBox["perfectdata", FontWeight->"Bold"], " with a total scaling and forced to be integer-valued" }], "Text", CellChangeTimes->{{3.4340333719075747`*^9, 3.4340333784453087`*^9}, { 3.4341455020840187`*^9, 3.4341456958872223`*^9}, {3.434145741305975*^9, 3.43414579629667*^9}, {3.4341458338177433`*^9, 3.434145855154063*^9}, { 3.434145902245934*^9, 3.4341459774687557`*^9}, {3.434148004531333*^9, 3.434148005818492*^9}, {3.434192275783101*^9, 3.434192283171342*^9}, { 3.434233367050811*^9, 3.434233368799025*^9}, {3.43423339928776*^9, 3.434233400151554*^9}}], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{ RowBox[{"listlength", "=", "10"}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"onetolength", " ", "=", " ", RowBox[{"Range", "[", "listlength", "]"}]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"thelifetime", " ", "=", " ", "50"}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{"dexp", "[", "x_", "]"}], ":=", " ", RowBox[{ FractionBox["1", "thelifetime"], " ", SuperscriptBox["\[ExponentialE]", FractionBox[ RowBox[{"-", " ", "x"}], "thelifetime"]]}]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"perfectdata", " ", "=", " ", RowBox[{"Map", "[", RowBox[{"dexp", ",", "onetolength"}], "]"}]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"mydata", " ", "=", " ", RowBox[{"Floor", "[", RowBox[{"1000", "perfectdata"}], "]"}]}], " ", ";"}], "\[IndentingNewLine]", RowBox[{"ListPlot", "[", "mydata", "]"}]}], "Input", CellChangeTimes->{{3.434033381766774*^9, 3.4340333991861773`*^9}, { 3.434145487108169*^9, 3.434145487304503*^9}, {3.434145613166511*^9, 3.434145651096015*^9}, {3.434145736729205*^9, 3.4341457369461737`*^9}, { 3.434145803099917*^9, 3.4341458324768553`*^9}, {3.434145993107635*^9, 3.4341459952388153`*^9}, {3.434146056529163*^9, 3.434146059228787*^9}, { 3.4341472689613132`*^9, 3.4341472745939903`*^9}, {3.4341473737233267`*^9, 3.434147374369298*^9}, {3.434147763652267*^9, 3.434147783361562*^9}, { 3.434147886276799*^9, 3.434147887980057*^9}, {3.434147919719145*^9, 3.4341479207384567`*^9}, {3.434147995014717*^9, 3.434148014270206*^9}, { 3.434148590140403*^9, 3.434148591216545*^9}, {3.4341916706373243`*^9, 3.4341916750915337`*^9}, {3.4341917184157953`*^9, 3.434191746149259*^9}, { 3.4341921336230917`*^9, 3.434192180163616*^9}, 3.434192600557564*^9, { 3.434199144634264*^9, 3.434199145022661*^9}, {3.434212913640128*^9, 3.4342129137275267`*^9}, {3.434229925308447*^9, 3.434229967138769*^9}, { 3.434233378537429*^9, 3.434233392697867*^9}}], Cell[BoxData[ GraphicsBox[ {Hue[0.67, 0.6, 0.6], PointBox[{{1., 19.}, {2., 19.}, {3., 18.}, {4., 18.}, {5., 18.}, {6., 17.}, {7., 17.}, {8., 17.}, {9., 16.}, {10., 16.}}]}, AspectRatio->NCache[GoldenRatio^(-1), 0.6180339887498948], Axes->True, AxesOrigin->{0, Automatic}, PlotRange->Automatic, PlotRangeClipping->True]], "Output", CellChangeTimes->{3.4342306671036787`*^9, 3.4342389546041937`*^9, 3.434240533883767*^9, 3.434241575240226*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"singlescounts", " ", RowBox[{ RowBox[{"PDF", "[", RowBox[{ RowBox[{"PoissonDistribution", "[", RowBox[{"rate", " ", "20", " ", SuperscriptBox["10", RowBox[{"-", "6"}]]}], "]"}], ",", "2"}], "]"}], "/", "250"}]}]], "Input", CellChangeTimes->{{3.433971294602928*^9, 3.433971375715934*^9}, { 3.4339714097190027`*^9, 3.433971446075726*^9}}], Cell[BoxData["2.8999999289799416`"], "Output", CellChangeTimes->{{3.433971364132535*^9, 3.433971446485813*^9}, 3.434024822754917*^9, 3.4342387077125473`*^9, 3.434240536531253*^9}] }, Open ]], Cell["\<\ So the naive expectation is that each bin will have 2.9 background events. \ This looks a bit high given my data, but I guess I'll work with this for a \ while. Anyway, the maximum likelihood (ML) analysis should catch the right \ parameter.\ \>", "Text", CellChangeTimes->{{3.433971457523335*^9, 3.433971542506768*^9}}] }, Closed]], Cell[CellGroupData[{ Cell["Parenthetical Note: total number of counts", "Subsection", CellChangeTimes->{{3.433967380309967*^9, 3.4339674034230423`*^9}}], Cell["\<\ The below is just a parenthetical note about how to get the total number of \ counts. \ \>", "Text", CellChangeTimes->{{3.4339673764186497`*^9, 3.433967394654971*^9}}], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{"totalcounts", " ", "=", " ", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"i", "=", "1"}], RowBox[{"Length", "[", "data", "]"}]], RowBox[{ RowBox[{"data", "[", RowBox[{"[", "i", "]"}], "]"}], "[", RowBox[{"[", "2", "]"}], "]"}]}]}], "\[IndentingNewLine]", RowBox[{"totalcounts", " ", "=", " ", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"i", "=", "1"}], RowBox[{"Length", "[", "data1", "]"}]], RowBox[{"data1", "[", RowBox[{"[", "i", "]"}], "]"}]}]}]}], "Input", CellChangeTimes->{{3.43396713900218*^9, 3.433967156047855*^9}, { 3.433967202739292*^9, 3.43396723744947*^9}, {3.4339673039309807`*^9, 3.433967307856204*^9}, {3.434240417051077*^9, 3.4342404321778517`*^9}}], Cell[BoxData["49128"], "Output", CellChangeTimes->{{3.4339672083140163`*^9, 3.433967238499689*^9}, 3.433967308619996*^9, 3.4340248249118757`*^9, 3.434238709616693*^9, { 3.434240411873115*^9, 3.434240432610791*^9}, 3.434240538260929*^9}], Cell[BoxData["49128"], "Output", CellChangeTimes->{{3.4339672083140163`*^9, 3.433967238499689*^9}, 3.433967308619996*^9, 3.4340248249118757`*^9, 3.434238709616693*^9, { 3.434240411873115*^9, 3.434240432610791*^9}, 3.434240538267248*^9}] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Construction of Likelihood and Log-Likelihood function: 1 parameter\ \>", "Subsection", CellChangeTimes->{{3.433971912667859*^9, 3.433971929835099*^9}, { 3.433971971980173*^9, 3.433971974236102*^9}}], Cell[TextData[{ "Let's start with a simple example of a one parameter fit. This is just ot \ get a feel for using maximum likelihood. We assume: no background, one fit, \ effectively infinite detector. Steps...\n1. ", StyleBox["Construct the likelihood function", FontWeight->"Bold"], ". Define the distribution function and evaluate the product of N such \ functions according to the observed data. It might be easier to start with \ the log of the likelihood function since this is relatively straightforward \ to do analytically.\n2. ", StyleBox["Find the maximum of this function", FontWeight->"Bold"], ". This may be nontrivial, but I can limit my search to points near where I \ expect to find the maximum. I should be able to use ", StyleBox["Mathematica", FontSlant->"Italic"], "'s built in optimize function, or otherwize I can just do a stupid scan.\n\ 3. 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To \ really get a good approximation I should repeat this procedure until the \ values of the parameters are stable. \n\nConcerns:\n1. I should be VERY \ careful with which value is \[Tau]p and which is \[Tau]m (tau-plus and \ tau-minus). In my definition of the exponential distribution I used one order \ (\[Tau]m, \[Tau]p), but in my SLog Likelihood I used the other order \ (\[Tau]p,\[Tau]m).\n2. For my write up: I should say a few words about the \ set up: what kind of scintillator, approximate volume?\n\nNotes: things to do \ next... \n1a. Try using ", StyleBox["Mathematica", FontSlant->"Italic"], "'s optimization functions for ", StyleBox["MySBinLikelihood", FontWeight->"Bold"], ".\n1b. Otherwise 'optimize' by hand by iterating the above procedure a few \ times until it's stable. That is to say I use the fit values and then vary \ each parameter over again.\n2. Try doing this without relying on Stirling's \ approximation. (Since Stirling's holds well for large N and since there's \ lots of data even in the large time bins, this shouldn't change \ appreciably.)" }], "Text", CellChangeTimes->{{3.434376845212358*^9, 3.434376869349712*^9}, { 3.434377544861663*^9, 3.434377653434194*^9}, {3.434377781674713*^9, 3.434377784722467*^9}, {3.43437791089349*^9, 3.434377996196001*^9}, { 3.434378064436686*^9, 3.434378229190712*^9}, {3.4343782715027227`*^9, 3.434378272067992*^9}, {3.4343800780009604`*^9, 3.434380099380666*^9}}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Second Go... 4 Parameter fit", "Subsection", CellChangeTimes->{{3.434415329193528*^9, 3.4344153351870127`*^9}}], Cell["\<\ So apparently I was doing something wrong above since I should be doing a \ FOUR parameter fit. 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The most reasonable values are tm = 15.92, tp = 24.22, bg = \ 18.75, tc = 62247. 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