Let $U$ and $V$ be independent random variables both having Bernoulli , Compute $E(W)$

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Let $U$ and $V$ be independent random variables both having Bernoulli distribution, i.e. $U$ ~ Bernoulli($0.2$) and $V$ ~ Bernoulli($0.2$)



Let $W = U cdot V$



(a) Compute $E(W)$



(b) Compute $E(UW)$



My attempt:



$(a)$



$$E(W) = E(UV) = E(U)E(V)$$



Because they're independent



$E(U) = E(V) = 0.2$, so $E(W) = 0.04$



$(b)$



$E(UW) = E(UUW) = E(U^2V) = E(U^2V) = E(U^2)E(V)$



$E(U^2) = 0.2^2 = 0.04$ [Not sure about this]



Thus,



$E(UW) = 0.04 cdot 0.2 = 0.008$




probability expectation




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  • 1




    Note $U^2=U$ here so $E[U^2]=E[U]$ not $left(E[U]right)^2$
    – Henry
    Jul 25 at 23:53















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Let $U$ and $V$ be independent random variables both having Bernoulli distribution, i.e. $U$ ~ Bernoulli($0.2$) and $V$ ~ Bernoulli($0.2$)



Let $W = U cdot V$



(a) Compute $E(W)$



(b) Compute $E(UW)$



My attempt:



$(a)$



$$E(W) = E(UV) = E(U)E(V)$$



Because they're independent



$E(U) = E(V) = 0.2$, so $E(W) = 0.04$



$(b)$



$E(UW) = E(UUW) = E(U^2V) = E(U^2V) = E(U^2)E(V)$



$E(U^2) = 0.2^2 = 0.04$ [Not sure about this]



Thus,



$E(UW) = 0.04 cdot 0.2 = 0.008$




probability expectation




share|cite|improve this question

















  • 1




    Note $U^2=U$ here so $E[U^2]=E[U]$ not $left(E[U]right)^2$
    – Henry
    Jul 25 at 23:53













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $U$ and $V$ be independent random variables both having Bernoulli distribution, i.e. $U$ ~ Bernoulli($0.2$) and $V$ ~ Bernoulli($0.2$)



Let $W = U cdot V$



(a) Compute $E(W)$



(b) Compute $E(UW)$



My attempt:



$(a)$



$$E(W) = E(UV) = E(U)E(V)$$



Because they're independent



$E(U) = E(V) = 0.2$, so $E(W) = 0.04$



$(b)$



$E(UW) = E(UUW) = E(U^2V) = E(U^2V) = E(U^2)E(V)$



$E(U^2) = 0.2^2 = 0.04$ [Not sure about this]



Thus,



$E(UW) = 0.04 cdot 0.2 = 0.008$




probability expectation




share|cite|improve this question













Let $U$ and $V$ be independent random variables both having Bernoulli distribution, i.e. $U$ ~ Bernoulli($0.2$) and $V$ ~ Bernoulli($0.2$)



Let $W = U cdot V$



(a) Compute $E(W)$



(b) Compute $E(UW)$



My attempt:



$(a)$



$$E(W) = E(UV) = E(U)E(V)$$



Because they're independent



$E(U) = E(V) = 0.2$, so $E(W) = 0.04$



$(b)$



$E(UW) = E(UUW) = E(U^2V) = E(U^2V) = E(U^2)E(V)$



$E(U^2) = 0.2^2 = 0.04$ [Not sure about this]



Thus,



$E(UW) = 0.04 cdot 0.2 = 0.008$





probability expectation





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share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 2:55









Graham Kemp

80.1k43275




80.1k43275









asked Jul 25 at 23:45









Tinler

404311




404311







  • 1




    Note $U^2=U$ here so $E[U^2]=E[U]$ not $left(E[U]right)^2$
    – Henry
    Jul 25 at 23:53













  • 1




    Note $U^2=U$ here so $E[U^2]=E[U]$ not $left(E[U]right)^2$
    – Henry
    Jul 25 at 23:53








1




1




Note $U^2=U$ here so $E[U^2]=E[U]$ not $left(E[U]right)^2$
– Henry
Jul 25 at 23:53





Note $U^2=U$ here so $E[U^2]=E[U]$ not $left(E[U]right)^2$
– Henry
Jul 25 at 23:53











2 Answers
2






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Note that



beginalign
E(U^2) & = 0 cdot P(U^2 = 0) + 1 cdot P(U^2 = 1) \
& = 0 cdot P(U = 0) + 1 cdot P(U = 1) qquad leftarrow textbecause $U$ is either $0$ or $1$ \
& = P(U = 1)
endalign






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    $mathsf E(UV) = mathsf E(U)mathsf E(V) = 0.2^2$ Yes.



    $mathsf E(U^2V) = mathsf E(U^2)mathsf E(V)$ Also.



    However, $requirecancelcancel~~mathsf E(U^2)=mathsf E(U)^2~~$ No! Rather: $mathsf E(U^2)=mathsfVar(U)+mathsf E(U)^2$



    Though, because $U$ is Bernouli, we can also use: $mathsf E(U^2)=mathsf P(U^2=1)=mathsf P(U=1)=mathsf E(U)$.






    share|cite|improve this answer























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      2 Answers
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      active

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      2 Answers
      2






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      up vote
      0
      down vote













      Note that



      beginalign
      E(U^2) & = 0 cdot P(U^2 = 0) + 1 cdot P(U^2 = 1) \
      & = 0 cdot P(U = 0) + 1 cdot P(U = 1) qquad leftarrow textbecause $U$ is either $0$ or $1$ \
      & = P(U = 1)
      endalign






      share|cite|improve this answer

























        up vote
        0
        down vote













        Note that



        beginalign
        E(U^2) & = 0 cdot P(U^2 = 0) + 1 cdot P(U^2 = 1) \
        & = 0 cdot P(U = 0) + 1 cdot P(U = 1) qquad leftarrow textbecause $U$ is either $0$ or $1$ \
        & = P(U = 1)
        endalign






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Note that



          beginalign
          E(U^2) & = 0 cdot P(U^2 = 0) + 1 cdot P(U^2 = 1) \
          & = 0 cdot P(U = 0) + 1 cdot P(U = 1) qquad leftarrow textbecause $U$ is either $0$ or $1$ \
          & = P(U = 1)
          endalign






          share|cite|improve this answer













          Note that



          beginalign
          E(U^2) & = 0 cdot P(U^2 = 0) + 1 cdot P(U^2 = 1) \
          & = 0 cdot P(U = 0) + 1 cdot P(U = 1) qquad leftarrow textbecause $U$ is either $0$ or $1$ \
          & = P(U = 1)
          endalign







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 25 at 23:50









          Brian Tung

          25.2k32353




          25.2k32353




















              up vote
              0
              down vote













              $mathsf E(UV) = mathsf E(U)mathsf E(V) = 0.2^2$ Yes.



              $mathsf E(U^2V) = mathsf E(U^2)mathsf E(V)$ Also.



              However, $requirecancelcancel~~mathsf E(U^2)=mathsf E(U)^2~~$ No! Rather: $mathsf E(U^2)=mathsfVar(U)+mathsf E(U)^2$



              Though, because $U$ is Bernouli, we can also use: $mathsf E(U^2)=mathsf P(U^2=1)=mathsf P(U=1)=mathsf E(U)$.






              share|cite|improve this answer



























                up vote
                0
                down vote













                $mathsf E(UV) = mathsf E(U)mathsf E(V) = 0.2^2$ Yes.



                $mathsf E(U^2V) = mathsf E(U^2)mathsf E(V)$ Also.



                However, $requirecancelcancel~~mathsf E(U^2)=mathsf E(U)^2~~$ No! Rather: $mathsf E(U^2)=mathsfVar(U)+mathsf E(U)^2$



                Though, because $U$ is Bernouli, we can also use: $mathsf E(U^2)=mathsf P(U^2=1)=mathsf P(U=1)=mathsf E(U)$.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  $mathsf E(UV) = mathsf E(U)mathsf E(V) = 0.2^2$ Yes.



                  $mathsf E(U^2V) = mathsf E(U^2)mathsf E(V)$ Also.



                  However, $requirecancelcancel~~mathsf E(U^2)=mathsf E(U)^2~~$ No! Rather: $mathsf E(U^2)=mathsfVar(U)+mathsf E(U)^2$



                  Though, because $U$ is Bernouli, we can also use: $mathsf E(U^2)=mathsf P(U^2=1)=mathsf P(U=1)=mathsf E(U)$.






                  share|cite|improve this answer















                  $mathsf E(UV) = mathsf E(U)mathsf E(V) = 0.2^2$ Yes.



                  $mathsf E(U^2V) = mathsf E(U^2)mathsf E(V)$ Also.



                  However, $requirecancelcancel~~mathsf E(U^2)=mathsf E(U)^2~~$ No! Rather: $mathsf E(U^2)=mathsfVar(U)+mathsf E(U)^2$



                  Though, because $U$ is Bernouli, we can also use: $mathsf E(U^2)=mathsf P(U^2=1)=mathsf P(U=1)=mathsf E(U)$.







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 26 at 2:55


























                  answered Jul 25 at 23:50









                  Graham Kemp

                  80.1k43275




                  80.1k43275






















                       

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