Tossing a triple of coins












1












$begingroup$


There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14


















1












$begingroup$


There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14
















1












1








1





$begingroup$


There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability







share|cite|improve this question











$endgroup$




There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 12:33







bard

















asked Nov 27 '18 at 10:40









bardbard

1886




1886








  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14
















  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14










1




1




$begingroup$
Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
$endgroup$
– nicola
Nov 27 '18 at 11:16






$begingroup$
Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
$endgroup$
– nicola
Nov 27 '18 at 11:16






1




1




$begingroup$
The variance Var(T1) above is the variance of the expectation, not the total variance.
$endgroup$
– KRKirov
Nov 29 '18 at 19:14






$begingroup$
The variance Var(T1) above is the variance of the expectation, not the total variance.
$endgroup$
– KRKirov
Nov 29 '18 at 19:14












1 Answer
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$begingroup$

The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



By the law of total variance,
$$begin{eqnarray}
Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
\ &=& mathbb{E}[P(1-P)] + Var(P)
\ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
\ &=& 0.5 - 0.5^2
\ &=& 0.25
end{eqnarray}$$



$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& 300 cdot Var(T_i)
\ &=& 300 cdot 0.25
\ &=& 75
end{eqnarray}$$



Still don't understand why my first approach is wrong though.






share|cite|improve this answer









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    $begingroup$

    The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



    By the law of total variance,
    $$begin{eqnarray}
    Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
    \ &=& mathbb{E}[P(1-P)] + Var(P)
    \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
    \ &=& 0.5 - 0.5^2
    \ &=& 0.25
    end{eqnarray}$$



    $$begin{eqnarray}
    Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
    \ &=& 300 cdot Var(T_i)
    \ &=& 300 cdot 0.25
    \ &=& 75
    end{eqnarray}$$



    Still don't understand why my first approach is wrong though.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



      By the law of total variance,
      $$begin{eqnarray}
      Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
      \ &=& mathbb{E}[P(1-P)] + Var(P)
      \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
      \ &=& 0.5 - 0.5^2
      \ &=& 0.25
      end{eqnarray}$$



      $$begin{eqnarray}
      Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
      \ &=& 300 cdot Var(T_i)
      \ &=& 300 cdot 0.25
      \ &=& 75
      end{eqnarray}$$



      Still don't understand why my first approach is wrong though.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



        By the law of total variance,
        $$begin{eqnarray}
        Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
        \ &=& mathbb{E}[P(1-P)] + Var(P)
        \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
        \ &=& 0.5 - 0.5^2
        \ &=& 0.25
        end{eqnarray}$$



        $$begin{eqnarray}
        Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
        \ &=& 300 cdot Var(T_i)
        \ &=& 300 cdot 0.25
        \ &=& 75
        end{eqnarray}$$



        Still don't understand why my first approach is wrong though.






        share|cite|improve this answer









        $endgroup$



        The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



        By the law of total variance,
        $$begin{eqnarray}
        Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
        \ &=& mathbb{E}[P(1-P)] + Var(P)
        \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
        \ &=& 0.5 - 0.5^2
        \ &=& 0.25
        end{eqnarray}$$



        $$begin{eqnarray}
        Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
        \ &=& 300 cdot Var(T_i)
        \ &=& 300 cdot 0.25
        \ &=& 75
        end{eqnarray}$$



        Still don't understand why my first approach is wrong though.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 12:57









        bardbard

        1886




        1886






























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