Hi Vera,
I reply to a couple of points, Daniele will reply to the first part
> I am still not quite sure what is done with the first bin:
> we add all data and MC up to the Mx cut into one bin, ok?
> but is that bin then treated in the fit un the same way as all other bins?
>
yes, C_u,C_c and C_o are extracted from a single fit to all bins
> In looking at all this, I am coming back to the earlier suggestion that we should
> perform a combined fit to the enhanced and depleted samples. The depleted
> sample should fix the b > c background, the enriched sample to extract the b> u
> signal. One would need to be careful to treat the errors and correlations correctly.
> Experienced users of MINUIT could assist here.
> This should allow us to understand the correlations between C_u and C_c!
> This should also help to estimate and limit the uncertainty on the s.l. branching ratios by checking the fit quality for different assumptions on the BR, not just the change in fit values.
>
this would introduce a dependency on the knowledge of the kaonID
efficiencies and misidentification rates which would be much larger than
the SL branching fraction (and worse determined)
> BTW: On page 62, Eq. 23 the parameters have incorrect subscripts,
> also, it should be made clear which interval in M_x is actually fitted.
> The statement about the signal events N_u=... is fuzzy as to the use
> and its relation to C_s!!
there is no relation to C_u (sorry for the incorrect labelling will be
fixed in the next version), the signal normalization is floated in the
fit, but then ignored. The formula for N_u is the one actually used.
> Likewise, Eq. 26 and the reasoning about the other backgrounds remains very obscure! \
> If Eq is exact as you state (which it is not), then please explain
> the rest more clearly. This is best done to write the exact formulation,
> and then show the simplification you chose. I tried this in the earlier version, but
> apparently I did not agree with Riccardo's view of this problem.
>
My point is that instead of the signal and the background measured
separately we use the linear combination in equation 26 because it is less
sensitive to the effect if mixing, but then we use the same formula to
calculate efficiencies and to estimate signal and background shapes.
> On a totally different subject: Figure 6b
> Shouldn't we expect to see a peaking here from this cross feed?
> Valerie apparently sees a small peak leading to a correction!?
this is done with mo lepton requirement, therefore also the peaking
background is covered by background. At the end of the selection there is
some peaking background and you can find its Mx distribution in
figure 90.
bye and Happy New Year
Ric
>
> Thanks again for input,
> Vera
>
>
>
>
>
>
>
>
> Original Message
> From: Del Re, Daniele
> Sent: Thursday, January 02, 2003 11:57 AM
> To: vubrecoil
> Subject: fitted BR for b>clnu events
>
>
>
> Hi,
>
> first of all happy New Year.
>
>
> As requested by Vera, I produced the results for the fitted
> BR for b>clnu events.
>
> Here you can find all numbers for the different subsamples.
> The measured values are still ratios of branching ratios, but now
> the expected result is 1.
>
> Actually it is not equal to 1 but to 1  ratio(BR)_bulnu (since
> in the numerator we have the b>clnu component while in the
> denominator we have the total semileptonic BR). Then the expected
> value is ~0.98.
>
> enriched sample
> 
>
> All BRBR = 0.931841 + 0.055113(stat)
>
> B0 BRBR = 0.931948 + 0.101485(stat)
> Bch BRBR = 0.949709 + 0.0652322(stat)
> ele BRBR = 0.914472 + 0.0692768(stat)
> mu BRBR = 0.917701 + 0.0837971(stat)
> run1 BRBR = 1.0263 + 0.0952589(stat)
> run2 BRBR = 0.890657 + 0.0607272(stat)
> sb1 BRBR = 1.01698 + 0.103764(stat)
> sb2 BRBR = 0.985293 + 0.0750252(stat)
> sb3 BRBR = 0.86291 + 0.0714007(stat)
>
>
> depleted sample
> 
>
> All BRBR = 0.947103 + 0.0280162(stat)
>
> B0 BRBR = 0.944411 + 0.0562612(stat)
> Bch BRBR = 0.965159 + 0.0328286(stat)
> ele BRBR = 0.9542 + 0.0375441(stat)
> mu BRBR = 0.923164 + 0.0381606(stat)
> run1 BRBR = 1.01548 + 0.0475409(stat)
> run2 BRBR = 0.920063 + 0.030727(stat)
> sb1 BRBR = 0.968563 + 0.060525(stat)
> sb2 BRBR = 0.998408 + 0.0395365(stat)
> sb3 BRBR = 0.909976 + 0.0380277(stat)
>
>
>
> I did not calculate the error due to the MC statistics but it
> should be of the same order of magnitude of the statistical one.
>
>
> These results are in agreement with the expected values for all the
> subsamples. This implies that the efficiencies on the background
> are well estimated.
>
>
> Daniele
>
>
