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STD Deviation using Partial Derivatives Method

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  • #52086

    Dona
    Participant

    I am given the task to compute the STD deviation of Y= 2*X1+3*X2^2, with X1=2, X1=3, std dev X1 = 0.1, std dev X2 = 0.2. I must use the partial dericatives method to get STD dev of Y. Here is my logic…I am not asking for a solution just some feed back on where I may be going wroing with this:
    dY/dX1 = 2+3*X2^2 and dY/dX2 = 2*X1+6*X2 substituting in X1 and X2 I get 29 and 22 respectively.
    Now if I want to find STD Dev of Y I need to use the following eqn:
    STD dev Y = SQRT {(29*0.1)^2 + (22*0.2)^2)} this yields 5.26
    Unfortunately 5.26 does not even come close to the choices I have for an answer. I have looked and looked and again and I am completely stumped on this one. I would appreciate any direction you could give me.

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    #182655

    clb1
    Participant

      The partial derivatives of the expression  Y = 2*X1 +3*X2^2  with respect to X1 and X2 are:
     dY/dX1 = 2
     dY/dX2 = 6*X2
     so if we plug these values into the equation you are using for the standard deviation of Y this would be
      sqrt( (2*.1)^2 +(6*3*.2)^2) = 3.606

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    #182659

    Mikel
    Member

    This thread is a joke, right?

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    #182663

    Bower Chiel
    Participant

    Hi Jon
    Presumably you mean that E[X1]=2 and E[X2]=3 since if X1=2 and X2=3 (I presume you meant to type X2) then Y would be a constant and would have variance 0!
    I’ve not seen partial derivatives used in this context so I thought: – Let’s try looking at the simpler case of Z=2X1+3X2. The partial derivatives of Z with respect to X1 and X2 are respectively 2 and 3 so your formula would appear to give the st. dev. of Z to be SQRT{(2*0.1)^2+(3*0.2)^2}which is exactly what you get from the widely known result for the standard deviation of a linear combination of two INDEPENDENT random variables. (You made no mention of independence.) Thus the partial derivative approach certainly works here.Next I used Minitab to simulate samples of 100000 values of X1 and X2 values (independent), calculated the 100000 values of Y and got st. dev. of 3.597, which is close to the value given by another poster. Thus the partial derivative theory appears to work here also. Is there a text-book associated with your course that includes the method involving partial derivatives? I’d be very interested to see a precise statement of the theory that you are being asked to apply.Best WishesBower Chiel

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    #182664

    Dona
    Participant

    Wow guys…this is a bit embarassing. I guess I need to refresh my memory on simple partial derivatives. Thanks for the kick in the right direction. I would be interested to see in detail how you solved this in Minitab though. Meaning which functions you used.                                                                                                                                                                         

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    #182670

    Bower Chiel
    Participant

    Hi JonYou can drive Minitab using commands after using Editor > Enable Commands. You type in commands after the MTB > prompt in the Session Window and you end a line with a ; if a subcommand is required. Hitting the enter key activates the command(s). My simulation is as follows: –
    MTB > Name c1 “x1”
    MTB > Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Enable Commands. You type in commands after the MTB > prompt in the Session Window and you end a line with a ; if a subcommand is required. Hitting the enter key activates the command(s). My simulation is as follows: –
    MTB > Name c1 “x1”
    MTB > Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Enable Commands. You type in commands after the MTB > prompt in the Session Window and you end a line with a ; if a subcommand is required. Hitting the enter key activates the command(s). My simulation is as follows: –
    MTB > Name c1 “x1”
    MTB > Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y prompt in the Session Window and you end a line with a ; if a subcommand is required. Hitting the enter key activates the command(s). My simulation is as follows: –
    MTB > Name c1 “x1”
    MTB > Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y prompt in the Session Window and you end a line with a ; if a subcommand is required. Hitting the enter key activates the command(s). My simulation is as follows: –
    MTB > Name c1 “x1”
    MTB > Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Name c1 “x1”
    MTB > Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Name c1 “x1”
    MTB > Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Random 100000 ‘x1’;
    SUBC> Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Normal 2 0.1.
    MTB > Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Name c2 “x2”
    MTB > Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Random 100000 ‘x2’;
    SUBC> Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Normal 3 0.2.
    MTB > Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Name C3 ‘y’
    MTB > Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB > Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Describe ‘y’;
    SUBC> Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Mean;
    SUBC> StDeviation;
    SUBC> Count.

    Descriptive Statistics: y StDeviation;
    SUBC> Count.

    Descriptive Statistics: y StDeviation;
    SUBC> Count.

    Descriptive Statistics: y Count.

    Descriptive Statistics: y Count.

    Descriptive Statistics: y Total
    Variable Count Mean StDev
    y 100000 31.147 3.597.I actually drove it via the menus but thought that commands would be easier to communicate to you. Note that what I’ve done does not PROVE anything!I use such simulations a lot in teaching – you can explain lots of things without having to delve into theory. Speaking of which you have not provided any reference to the theory you are applying – could you please do so?Best WishesBower Chiel

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    #182672

    Bower Chiel
    Participant

    Hi JonApologies for the dog’s breakfast! The commands somehow have been repeated over and over again. Please post an e-mail address and I’ll send you a copy of the Minitab project.Best WishesBower Chiel

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    #182681

    Dona
    Participant

    Here is my email address…[email protected]
    As fo rthe theory I do not have any written instructions. This was a verbal commnuication of the problem. Sorry. I wish I had soemthing to go from.
     

    0
    #182687

    BTDT
    Participant

    BC:The “greater than” symbol confuses the html interpreter on the forum interface. Could you repost using a comma or something? This looks really interesting.Cheers, Alastair

    0
    #182688

    Bower Chiel
    Participant

    Hi Jon, BTDTTake 2! If this one appears as a dog’s breakfast then post an e-mail address and I’ll happily forward the Minitab Project file. You can drive Minitab using commands after using Editor followed by Enable Commands. You type in commands after the MTB prompt in the Session Window. You end a line with a ; (semi-colon) if a subcommand is required and a SUBC prompt appears. A stop sign is required after the last subcommand used after any MTB prompt. The MTB and SUBC prompts appear with a greater than sign after them on screen but BTDT has kindly explained that typing the actual symbol created the dog’s breakfast mess in a previous post! Hitting the enter key activates the command(s).My simulation is as follows: –

    MTB Name c1 “x1”
    MTB Random 100000 ‘x1’;
    SUBC Normal 2 0.1.
    MTB Name c2 “x2”
    MTB Random 100000 ‘x2’;
    SUBC Normal 3 0.2.
    MTB Name C3 ‘y’
    MTB Let ‘y’ = 2*’x1’+3*’x2’**2
    MTB Describe ‘y’;
    SUBC Mean;
    SUBC StDeviation;
    SUBC> Count. Count. Count.Descriptive Statistics: y
    Variable Count Mean StDev
    y 100000 31.147 3.597.I actually drove it via the menus but thought that commands would be easier to communicate to you. Note that what I’ve done does not PROVE anything!Best WishesBower Chiel

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    #182698

    BTDT
    Participant

    Jon:Great post! I love forcing myself to relearn the old stuff.A good reference with an example can be found at:http://www.itl.nist.gov/div898/handbook/mpc/section6/mpc64.htmA few points for you:1) Review how you take partial derivatives of a function2) The Monte Carlo method is a good way of checking your answer3) The distribution of the error function follows a gamma distribution, with scale=beta, shape=alpha. The standard deviation of the gamma function is =scale*sqrt(beta). The final distributions are approximately normally distributed.4) The results are relatively independent of the type of distribution for X1 and X2The exact value using partial derivatives is 3.606 assuming zero correlation between X1 and X2.I performed a few Monte Carlo simulations using Crystal Ball.- 1,000,000 trials- 99.7% of the total variation is due to the variation in X2 alone- When the correlation between X1 and X2 is high (~0.95). The total variation is about 15% larger than when you assume X1 is independent of X2.- Results are independent of the distribution of X1 or X2.- The fits to the final distributions are- Normal in X1 and X2: alpha=55.46166, beta=0.47 (stdev=3.607)- Lognormal in X1 and X2: alpha=58.90143, beta=0.47 (stdev=3.607)Cheers, Alastair

    0
    #182699

    BTDT
    Participant

    BC:I repeated the simulation using Crystal Ball.https://www.isixsigma.com/forum/showmessage.asp?messageID=155162The assumption in the derivation is that the errors (variations) in X1 and X2 are independent: zero correlation. CB allows me to include correlations and change the distribution types. The results are the same as you achieved using Minitab.Cheers, Alastair

    0
    #182700

    clb1
    Participant

    BTDT,
      Old Stuff????  Just remember, every book is new until you have read it and I’d say the same has to be true of mathematical methods.  After providing the correct partial derivatives I was hoping Jon would tell us more about the standard deviation formula since I had never seen it. 
          Thanks for the additional information.

    0
    #182703

    BTDT
    Participant

    clb1:I mean old stuff because as I was researching this it brought together all kinds of things I hadn’t used in detail for over a decade (Taylor series, partial differentials, etc.). The link to “error propogation” was pretty helpful and I can see how it generalizes to the use of geometric mean for standard deviations for linear functions. Once I put the pieces together, it made perfect sense. I added the extra investigation in the effect of correlation. As correlations become high, the risk(variation) increases. I can see, in general terms, the holes in Charles Li’s methods for pricing collateralized debt obligations (CDOs) and credit defult swaps (CDS)s. His Gaussian copula function reduces all correlations to a single value (gamma). It is a simple way to not having to estimate all those correlations, but is a poor way of estimating the risk (variation) of a basket of CDOs.Lots of other studies have shown that the pricing models were far too sensitive to the hundreds of correlations that the method ignored.This is a round about way of saying that if investment bankers knew more about things, then we wouldn’t all be on a trillion dollar bailout hook.Cheers, Alastair

    0
    #182712

    Bower Chiel
    Participant

    Hi AllThanks to Alistair for providing the context. A bit of searching has revealed that W Edwards Deming dealt with this stuff in his 1966 book Some Theory of Sampling – equation 48 on p.130 – which you can read at http://books.google.co.uk/books?id=xQCvWLpAP-IC&pg=PA127&lpg=PA127&dq=propagation+of+error+theory&source=bl&ots=NZxEpXHdgb&sig=BzF_NbwEB8QWUA0ZPtaSttEWMAE&hl=en&ei=Q2LJSaOyEY-OjAf9jcjNAw&sa=X&oi=book_result&resnum=1&ct=result#PPA130,M1 .
    Best Wishes
    Bower Chiel

    0
    #182715

    eric
    Participant

    This is a very intersting post. I posted something similar to this a few days back but no one replied. I need to compute the DPMO givem Y.
    Y = 10 + 2*X1 +3*X2,
    with long term std dev Y = 0.1+0.2*X1*X2, where X1= 2, X2 = 2 & USL =18.
    I have been told I could use the partial derivatives method to solve this? So far it appears this problem is very similar to Jon’s.
     
    If I substitute X1 and X2 into Y eqn I get 20.
     
    STD DEVlt = 0.1 + 0.2*X1*X2,  so I sub in  X1 and X2 and get STD Devlt = 0.9, not sure if this is on the right track or not.
     
    Using partial derivatives method:
    dY/dX1 = 2
    dY/dX2 = 3
    STD Y = SQRT [(2*.9)^2 + (3*.9)^2] = 3.24
     
    Zusl = (18-20)/3.24= 0.617, when I find the corresponding Z value in the table this does not line up with any of the answers I have for choices.
     
    Any help would be appreciated
     
    CHEERS! 
           
     

    0
    #182716

    BTDT
    Participant

    Eric:I’ll have a look at this in a while – I got busy again.Cheers, Alastair

    0
    #182718

    Bower Chiel
    Participant

    Hi EricYour question does not appear to be precisely stated. If you have Y=10+2*X1+3*X2 then Y is a random variable which is a function of the two random variables X1 and X2. To find the mean and standard deviation of Y would require information on the means and standard deviations of X1 and X2 e.g. if X1 had mean 2 i.e. that the expected value of X1 is 2 or in short-hand E[X1]=2 and that also E[X2]=2. Then E[Y]=10+2*2+3*2=20 i.e. the mean of Y is 20. Suppose for illustration purposes that X1 had standard deviation 0.1 and that X2 had standard deviation 0.2. The partial diff formula would then give standard deviation of Y to be sqrt((2*0.1)^2+(3*0.2))=0.632. You could then move on to your capability analysis.
    You state that the formula 0.1+0.2*X1*X2 gives the standard deviation of Y so if that is really the case you cannot take the result from that formula as the standard deviation of both X1 and X2. Also if you have a formula for the standard deviation of Y you would not need to use the partial diff approach to work out something you’ve already got! Hope this makes some sense.
    Best Wishes
    Bower Chiel

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    #182723

    eric
    Participant

    Cheers…you are correct my assumtion is that X1=2 and X2 = 2 are the means. Therefore,the mean of Y is = 20 as you stated. The long term std deviation of Y is = 0.1+0.2*X1*X2. If that is truly the STD dev of Y can this be as simple as plugging in X1 & X2 into the long tern std dev eqn Y is = 0.1+0.2*X1*X2 = 0.9 then using the following formula Z = (USL – mean)/std dev to compute a Z value and calculate my DPMO?
    So with that said something here does not make sense to me…if my mean for Y = 20 and the USL is 18 with a std deviation for Y of 0.9. Just considering this form a graphical perspective it does not make sense. The mean should be lower than the USL….where am I going wrong here? Sorry I am a bit slow :)

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    #182735

    eric
    Participant

    I think I got a bit further on this problem. I can calculate Zst using the following:
    Zst = ((18-20)/.9)) +1.5 = -.720
    I found an excel sheet that I can use to calc DPMO which is using the follwing equation to do this:
    (1-NORMSDIST(-.720 -1.5))*1000000.
    I am getting the correct answer but am now bothered that I cannot seem to do this by hand. How do I convert from my Z value to DPMO?
    regards

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    #182797

    eric
    Participant

    Hello I am still a bit stumped on how to get DPMO…any help would be appreciated….thanks!

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