$X_{1},…,X_{n}$ ~ $Ber(p)$ what can I say about $exp(lambda X_{1}),…,exp(lambda X_{n})$












1












$begingroup$


Let $X_{1},...,X_{n}$ ~ $Ber(p)$ be independent, and $lambda > 0$. What can I say about $exp(lambda X_{1}),...,exp(lambda X_{n})$ with respect to $mathbb E[exp(lambda X_{i})]$?



Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.



I assume $exp(lambda X_{i})$ will be bernoulli distributed but I fail to find a way to determine the Bernoulli coefficients, and how to prove these. Any ideas?






probability-theory probability-distributions random-variables independence







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $X_{1},...,X_{n}$ ~ $Ber(p)$ be independent, and $lambda > 0$. What can I say about $exp(lambda X_{1}),...,exp(lambda X_{n})$ with respect to $mathbb E[exp(lambda X_{i})]$?



    Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.



    I assume $exp(lambda X_{i})$ will be bernoulli distributed but I fail to find a way to determine the Bernoulli coefficients, and how to prove these. Any ideas?






    probability-theory probability-distributions random-variables independence







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X_{1},...,X_{n}$ ~ $Ber(p)$ be independent, and $lambda > 0$. What can I say about $exp(lambda X_{1}),...,exp(lambda X_{n})$ with respect to $mathbb E[exp(lambda X_{i})]$?



      Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.



      I assume $exp(lambda X_{i})$ will be bernoulli distributed but I fail to find a way to determine the Bernoulli coefficients, and how to prove these. Any ideas?






      probability-theory probability-distributions random-variables independence







      share|cite|improve this question











      $endgroup$




      Let $X_{1},...,X_{n}$ ~ $Ber(p)$ be independent, and $lambda > 0$. What can I say about $exp(lambda X_{1}),...,exp(lambda X_{n})$ with respect to $mathbb E[exp(lambda X_{i})]$?



      Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.



      I assume $exp(lambda X_{i})$ will be bernoulli distributed but I fail to find a way to determine the Bernoulli coefficients, and how to prove these. Any ideas?






      probability-theory probability-distributions random-variables independence




      probability-theory probability-distributions random-variables independence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 13:12









      Ahmad Bazzi

      8,0362724




      8,0362724










      asked Jan 10 at 12:32









      SABOYSABOY

      588311




      588311






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$


          Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.




          This is not true. Since $X_k$ are all discrete, then so is $exp(lambda X_k)$. Using the rule of discrete expectation, we can say that
          $$E(exp(lambda X)) = sum_{x in lbrace 0,1 rbrace } Pr(X=x) exp(lambda x) = Pr(X=0)exp(lambda (0)) + Pr(X=1)exp(lambda (1))$$
          Since $Pr(X=0) = 1-p$ and $Pr(X=1) = p$, we get
          $$E(exp(lambda X)) = 1-p+ pexp(lambda )$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why is $(f_{lambda}(X_{i}))_{i=1,...n}$ not i.i.d.?
            $endgroup$
            – SABOY
            Jan 10 at 19:54












          • $begingroup$
            @SABOY i did not say it is not i.i.d .. i said it is not true meaning that it is not a continuous random variable. It is a discrete random variable.
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:56










          • $begingroup$
            Apologies, misunderstanding
            $endgroup$
            – SABOY
            Jan 10 at 19:57










          • $begingroup$
            no worries :) @SABOY
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:59



















          1












          $begingroup$

          Guide:



          If $X_i=0$, we have $exp(lambda X_i)=1$, this happens with probability $1-p$.



          What happens to $exp(lambda X_i)$ if $X_i=1$? what is the corresponding probability.



          Now use the definition of expectation to compute the desired quantity.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $X_{i}=1$ then $exp(lambda X_{i})=exp(lambda)$ The corresponding probability would be $0$ would it not, unless $lambda = 0$?
            $endgroup$
            – SABOY
            Jan 10 at 12:46












          • $begingroup$
            the correponding probability should be $p$.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 12:49










          • $begingroup$
            So we've got $P(exp(lambda X_{i})=exp(lambda))=p$ while $P(exp(lambda X_{i})=1)=1-p$, so a kind of Bernoulli distribution between two values $exp(lambda)$ and $1$?
            $endgroup$
            – SABOY
            Jan 10 at 13:01












          • $begingroup$
            It is a discrete distribution that only takes two values.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 13:06











          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%2f3068588%2fx-1-x-n-berp-what-can-i-say-about-exp-lambda-x-1-exp%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$


          Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.




          This is not true. Since $X_k$ are all discrete, then so is $exp(lambda X_k)$. Using the rule of discrete expectation, we can say that
          $$E(exp(lambda X)) = sum_{x in lbrace 0,1 rbrace } Pr(X=x) exp(lambda x) = Pr(X=0)exp(lambda (0)) + Pr(X=1)exp(lambda (1))$$
          Since $Pr(X=0) = 1-p$ and $Pr(X=1) = p$, we get
          $$E(exp(lambda X)) = 1-p+ pexp(lambda )$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why is $(f_{lambda}(X_{i}))_{i=1,...n}$ not i.i.d.?
            $endgroup$
            – SABOY
            Jan 10 at 19:54












          • $begingroup$
            @SABOY i did not say it is not i.i.d .. i said it is not true meaning that it is not a continuous random variable. It is a discrete random variable.
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:56










          • $begingroup$
            Apologies, misunderstanding
            $endgroup$
            – SABOY
            Jan 10 at 19:57










          • $begingroup$
            no worries :) @SABOY
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:59
















          1












          $begingroup$


          Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.




          This is not true. Since $X_k$ are all discrete, then so is $exp(lambda X_k)$. Using the rule of discrete expectation, we can say that
          $$E(exp(lambda X)) = sum_{x in lbrace 0,1 rbrace } Pr(X=x) exp(lambda x) = Pr(X=0)exp(lambda (0)) + Pr(X=1)exp(lambda (1))$$
          Since $Pr(X=0) = 1-p$ and $Pr(X=1) = p$, we get
          $$E(exp(lambda X)) = 1-p+ pexp(lambda )$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why is $(f_{lambda}(X_{i}))_{i=1,...n}$ not i.i.d.?
            $endgroup$
            – SABOY
            Jan 10 at 19:54












          • $begingroup$
            @SABOY i did not say it is not i.i.d .. i said it is not true meaning that it is not a continuous random variable. It is a discrete random variable.
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:56










          • $begingroup$
            Apologies, misunderstanding
            $endgroup$
            – SABOY
            Jan 10 at 19:57










          • $begingroup$
            no worries :) @SABOY
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:59














          1












          1








          1





          $begingroup$


          Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.




          This is not true. Since $X_k$ are all discrete, then so is $exp(lambda X_k)$. Using the rule of discrete expectation, we can say that
          $$E(exp(lambda X)) = sum_{x in lbrace 0,1 rbrace } Pr(X=x) exp(lambda x) = Pr(X=0)exp(lambda (0)) + Pr(X=1)exp(lambda (1))$$
          Since $Pr(X=0) = 1-p$ and $Pr(X=1) = p$, we get
          $$E(exp(lambda X)) = 1-p+ pexp(lambda )$$






          share|cite|improve this answer









          $endgroup$




          Well because $f_{lambda }(x):=e^{lambda x}$ is a continuous random variable, $(f_{lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.




          This is not true. Since $X_k$ are all discrete, then so is $exp(lambda X_k)$. Using the rule of discrete expectation, we can say that
          $$E(exp(lambda X)) = sum_{x in lbrace 0,1 rbrace } Pr(X=x) exp(lambda x) = Pr(X=0)exp(lambda (0)) + Pr(X=1)exp(lambda (1))$$
          Since $Pr(X=0) = 1-p$ and $Pr(X=1) = p$, we get
          $$E(exp(lambda X)) = 1-p+ pexp(lambda )$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 13:19









          Ahmad BazziAhmad Bazzi

          8,0362724




          8,0362724












          • $begingroup$
            Why is $(f_{lambda}(X_{i}))_{i=1,...n}$ not i.i.d.?
            $endgroup$
            – SABOY
            Jan 10 at 19:54












          • $begingroup$
            @SABOY i did not say it is not i.i.d .. i said it is not true meaning that it is not a continuous random variable. It is a discrete random variable.
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:56










          • $begingroup$
            Apologies, misunderstanding
            $endgroup$
            – SABOY
            Jan 10 at 19:57










          • $begingroup$
            no worries :) @SABOY
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:59


















          • $begingroup$
            Why is $(f_{lambda}(X_{i}))_{i=1,...n}$ not i.i.d.?
            $endgroup$
            – SABOY
            Jan 10 at 19:54












          • $begingroup$
            @SABOY i did not say it is not i.i.d .. i said it is not true meaning that it is not a continuous random variable. It is a discrete random variable.
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:56










          • $begingroup$
            Apologies, misunderstanding
            $endgroup$
            – SABOY
            Jan 10 at 19:57










          • $begingroup$
            no worries :) @SABOY
            $endgroup$
            – Ahmad Bazzi
            Jan 10 at 19:59
















          $begingroup$
          Why is $(f_{lambda}(X_{i}))_{i=1,...n}$ not i.i.d.?
          $endgroup$
          – SABOY
          Jan 10 at 19:54






          $begingroup$
          Why is $(f_{lambda}(X_{i}))_{i=1,...n}$ not i.i.d.?
          $endgroup$
          – SABOY
          Jan 10 at 19:54














          $begingroup$
          @SABOY i did not say it is not i.i.d .. i said it is not true meaning that it is not a continuous random variable. It is a discrete random variable.
          $endgroup$
          – Ahmad Bazzi
          Jan 10 at 19:56




          $begingroup$
          @SABOY i did not say it is not i.i.d .. i said it is not true meaning that it is not a continuous random variable. It is a discrete random variable.
          $endgroup$
          – Ahmad Bazzi
          Jan 10 at 19:56












          $begingroup$
          Apologies, misunderstanding
          $endgroup$
          – SABOY
          Jan 10 at 19:57




          $begingroup$
          Apologies, misunderstanding
          $endgroup$
          – SABOY
          Jan 10 at 19:57












          $begingroup$
          no worries :) @SABOY
          $endgroup$
          – Ahmad Bazzi
          Jan 10 at 19:59




          $begingroup$
          no worries :) @SABOY
          $endgroup$
          – Ahmad Bazzi
          Jan 10 at 19:59











          1












          $begingroup$

          Guide:



          If $X_i=0$, we have $exp(lambda X_i)=1$, this happens with probability $1-p$.



          What happens to $exp(lambda X_i)$ if $X_i=1$? what is the corresponding probability.



          Now use the definition of expectation to compute the desired quantity.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $X_{i}=1$ then $exp(lambda X_{i})=exp(lambda)$ The corresponding probability would be $0$ would it not, unless $lambda = 0$?
            $endgroup$
            – SABOY
            Jan 10 at 12:46












          • $begingroup$
            the correponding probability should be $p$.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 12:49










          • $begingroup$
            So we've got $P(exp(lambda X_{i})=exp(lambda))=p$ while $P(exp(lambda X_{i})=1)=1-p$, so a kind of Bernoulli distribution between two values $exp(lambda)$ and $1$?
            $endgroup$
            – SABOY
            Jan 10 at 13:01












          • $begingroup$
            It is a discrete distribution that only takes two values.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 13:06
















          1












          $begingroup$

          Guide:



          If $X_i=0$, we have $exp(lambda X_i)=1$, this happens with probability $1-p$.



          What happens to $exp(lambda X_i)$ if $X_i=1$? what is the corresponding probability.



          Now use the definition of expectation to compute the desired quantity.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $X_{i}=1$ then $exp(lambda X_{i})=exp(lambda)$ The corresponding probability would be $0$ would it not, unless $lambda = 0$?
            $endgroup$
            – SABOY
            Jan 10 at 12:46












          • $begingroup$
            the correponding probability should be $p$.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 12:49










          • $begingroup$
            So we've got $P(exp(lambda X_{i})=exp(lambda))=p$ while $P(exp(lambda X_{i})=1)=1-p$, so a kind of Bernoulli distribution between two values $exp(lambda)$ and $1$?
            $endgroup$
            – SABOY
            Jan 10 at 13:01












          • $begingroup$
            It is a discrete distribution that only takes two values.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 13:06














          1












          1








          1





          $begingroup$

          Guide:



          If $X_i=0$, we have $exp(lambda X_i)=1$, this happens with probability $1-p$.



          What happens to $exp(lambda X_i)$ if $X_i=1$? what is the corresponding probability.



          Now use the definition of expectation to compute the desired quantity.






          share|cite|improve this answer









          $endgroup$



          Guide:



          If $X_i=0$, we have $exp(lambda X_i)=1$, this happens with probability $1-p$.



          What happens to $exp(lambda X_i)$ if $X_i=1$? what is the corresponding probability.



          Now use the definition of expectation to compute the desired quantity.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 12:35









          Siong Thye GohSiong Thye Goh

          100k1466117




          100k1466117












          • $begingroup$
            If $X_{i}=1$ then $exp(lambda X_{i})=exp(lambda)$ The corresponding probability would be $0$ would it not, unless $lambda = 0$?
            $endgroup$
            – SABOY
            Jan 10 at 12:46












          • $begingroup$
            the correponding probability should be $p$.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 12:49










          • $begingroup$
            So we've got $P(exp(lambda X_{i})=exp(lambda))=p$ while $P(exp(lambda X_{i})=1)=1-p$, so a kind of Bernoulli distribution between two values $exp(lambda)$ and $1$?
            $endgroup$
            – SABOY
            Jan 10 at 13:01












          • $begingroup$
            It is a discrete distribution that only takes two values.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 13:06


















          • $begingroup$
            If $X_{i}=1$ then $exp(lambda X_{i})=exp(lambda)$ The corresponding probability would be $0$ would it not, unless $lambda = 0$?
            $endgroup$
            – SABOY
            Jan 10 at 12:46












          • $begingroup$
            the correponding probability should be $p$.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 12:49










          • $begingroup$
            So we've got $P(exp(lambda X_{i})=exp(lambda))=p$ while $P(exp(lambda X_{i})=1)=1-p$, so a kind of Bernoulli distribution between two values $exp(lambda)$ and $1$?
            $endgroup$
            – SABOY
            Jan 10 at 13:01












          • $begingroup$
            It is a discrete distribution that only takes two values.
            $endgroup$
            – Siong Thye Goh
            Jan 10 at 13:06
















          $begingroup$
          If $X_{i}=1$ then $exp(lambda X_{i})=exp(lambda)$ The corresponding probability would be $0$ would it not, unless $lambda = 0$?
          $endgroup$
          – SABOY
          Jan 10 at 12:46






          $begingroup$
          If $X_{i}=1$ then $exp(lambda X_{i})=exp(lambda)$ The corresponding probability would be $0$ would it not, unless $lambda = 0$?
          $endgroup$
          – SABOY
          Jan 10 at 12:46














          $begingroup$
          the correponding probability should be $p$.
          $endgroup$
          – Siong Thye Goh
          Jan 10 at 12:49




          $begingroup$
          the correponding probability should be $p$.
          $endgroup$
          – Siong Thye Goh
          Jan 10 at 12:49












          $begingroup$
          So we've got $P(exp(lambda X_{i})=exp(lambda))=p$ while $P(exp(lambda X_{i})=1)=1-p$, so a kind of Bernoulli distribution between two values $exp(lambda)$ and $1$?
          $endgroup$
          – SABOY
          Jan 10 at 13:01






          $begingroup$
          So we've got $P(exp(lambda X_{i})=exp(lambda))=p$ while $P(exp(lambda X_{i})=1)=1-p$, so a kind of Bernoulli distribution between two values $exp(lambda)$ and $1$?
          $endgroup$
          – SABOY
          Jan 10 at 13:01














          $begingroup$
          It is a discrete distribution that only takes two values.
          $endgroup$
          – Siong Thye Goh
          Jan 10 at 13:06




          $begingroup$
          It is a discrete distribution that only takes two values.
          $endgroup$
          – Siong Thye Goh
          Jan 10 at 13:06


















          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%2f3068588%2fx-1-x-n-berp-what-can-i-say-about-exp-lambda-x-1-exp%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

          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

          Antonio Litta Visconti Arese

          Image
          Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
          Read more