Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer












3












$begingroup$


A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.



(a) Describe the statistical model for the observed data ($T$)



(b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer



(b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer




My attempt:



(a) Since geometric is iid with Negative Binomial



Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.



The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$



parameter is $theta$ and parameter space is $[0,1]$



(b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure






probability statistics







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.



    (a) Describe the statistical model for the observed data ($T$)



    (b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer



    (b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer




    My attempt:



    (a) Since geometric is iid with Negative Binomial



    Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.



    The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$



    parameter is $theta$ and parameter space is $[0,1]$



    (b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure






    probability statistics







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.



      (a) Describe the statistical model for the observed data ($T$)



      (b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer



      (b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer




      My attempt:



      (a) Since geometric is iid with Negative Binomial



      Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.



      The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$



      parameter is $theta$ and parameter space is $[0,1]$



      (b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure






      probability statistics







      share|cite|improve this question











      $endgroup$




      A random sample of $6$ observations $(X_1, X_2, cdots, X_6)$ is generated from a Geometric($theta$), where $theta in (0, 1)$ unknown, but only $T = sum_{i=1}^{6} X_i$ is observed by the statistician.



      (a) Describe the statistical model for the observed data ($T$)



      (b)-(i) Is it possible to parameterize the model by $Psi = frac{1-theta}{theta}$ ? Prove your answer



      (b)-(ii) Is it possible to parameterize the model by $Psi = theta(1-theta)$ ? Prove your answer




      My attempt:



      (a) Since geometric is iid with Negative Binomial



      Each $X_i$ ~ $mathrm{Geometric}(theta)$ therefore $T = sum_{i=1}^{6}$ ~ $mathrm{NegativeBinomial}(r, theta)$ where $theta in (0, 1)$ unknown.



      The probability function for T is given by $$f_{theta}(t) = {t +r-1choose t}(1-theta)^t theta^r$$ for $t = 0,1, cdots, 6$



      parameter is $theta$ and parameter space is $[0,1]$



      (b)-(i) and (b)-(ii) I'm not sure how to do. Would I just show they are one to one by graphing each of them? Not sure






      probability statistics




      probability statistics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 25 at 4:04









      Lee David Chung Lin

      4,38031242




      4,38031242










      asked Jan 25 at 3:32









      Bas basBas bas

      49012




      49012






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$



          For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$



          For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
            $endgroup$
            – Bas bas
            Jan 25 at 13:37












          • $begingroup$
            A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
            $endgroup$
            – Henry
            Jan 25 at 14:30











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086668%2fis-it-possible-to-parameterize-the-model-by-psi-frac1-theta-theta-p%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$



          For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$



          For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
            $endgroup$
            – Bas bas
            Jan 25 at 13:37












          • $begingroup$
            A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
            $endgroup$
            – Henry
            Jan 25 at 14:30
















          0












          $begingroup$

          Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$



          For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$



          For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
            $endgroup$
            – Bas bas
            Jan 25 at 13:37












          • $begingroup$
            A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
            $endgroup$
            – Henry
            Jan 25 at 14:30














          0












          0








          0





          $begingroup$

          Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$



          For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$



          For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$






          share|cite|improve this answer









          $endgroup$



          Your answer for (a) is a probability mass function so personally I would not use $f$ to describe it as that is often used for a density. I would write $6$ rather than $r$. And it depends on whether your negative binomial starts at $0$ or $6$, i.e. whether your geometric distributions start at $0$ or $1$



          For (b)(i) you can say that $Psi = frac{1-theta}{theta} iff theta = frac{1}{Psi+1}$ at least if $theta not = 0$ which it has to be for the question to be sensible, so you can just use this as a substitution to parameterize the model by $Psi$



          For (b)(ii) you can say that $Psi = theta({1-theta})$ is satisfied by $theta = frac{1 pm sqrt{1+4Psi}}{2}$, and the $pm$ is a warning sign; for example $Psi=0.09$ could be produced by $theta=0.1$ or $theta=0.9$ which would give you different probabilities, so so you cannot parameterize the model by this $Psi$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 8:43









          HenryHenry

          101k481168




          101k481168












          • $begingroup$
            "And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
            $endgroup$
            – Bas bas
            Jan 25 at 13:37












          • $begingroup$
            A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
            $endgroup$
            – Henry
            Jan 25 at 14:30


















          • $begingroup$
            "And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
            $endgroup$
            – Bas bas
            Jan 25 at 13:37












          • $begingroup$
            A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
            $endgroup$
            – Henry
            Jan 25 at 14:30
















          $begingroup$
          "And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
          $endgroup$
          – Bas bas
          Jan 25 at 13:37






          $begingroup$
          "And it depends on whether your negative binomial starts at 0 or 6, i.e. whether your geometric distributions start at 0 or 1". What do you mean? Confuse at this part get the rest though
          $endgroup$
          – Bas bas
          Jan 25 at 13:37














          $begingroup$
          A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
          $endgroup$
          – Henry
          Jan 25 at 14:30




          $begingroup$
          A negative binomial distribution may count the number of failures before the $n$th success (minimum possible value is $0$) or the number of attempts when then $n$th success occurs (minimum possible value is $n$). When $n=1$ you have a geometric distribution with a minimum possible value of $0$ or $1$ depending on which definition you are using; your question does not say.
          $endgroup$
          – Henry
          Jan 25 at 14:30


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086668%2fis-it-possible-to-parameterize-the-model-by-psi-frac1-theta-theta-p%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
          Read more