$X_{1},…,X_{n}$ ~ $Ber(p)$ 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











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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          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


















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