Compute $ Var(X+Y+Z) $ where $ X,Y,Z sim Binomial $












0












$begingroup$


Suppose I throw 3 fair dice 30 times.



Let,



X = no' of throws in which we don't get 4

Y = no' of throws in which we get 4 in only one die (out of 3)

Z = no' of throws in which we get 4 in exactly two dice (out of 3)



Compute $ Var(X+Y+Z) $



by definition $ Var(X+Y+Z) = E[(X+Y+Z)^2] + (E[X+Y+Z])^2 $



$ E[X+Y+Z] = E[X] + E[Y] + E[Z] = 29.8611$ because each variable of a multinomial distribution is of binomial distribution. so I know $ (E[X+Y+Z])^2 = 29.8611^2 $ but how do I calculate $ E[(X+Y+Z)^2] $ ?



Suggestions please?






probability variance expected-value







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$endgroup$








  • 1




    $begingroup$
    Consider the probability of getting three $4$s and then consider $Var(n-X-Y-Z)$
    $endgroup$
    – Henry
    Jan 3 at 21:11








  • 1




    $begingroup$
    Can I just set new variable $ W = $ no' of throws in which I get up to two 4's and then calculate the variance of $ W $ as it is binomial variable?
    $endgroup$
    – bm1125
    Jan 4 at 12:49






  • 1




    $begingroup$
    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    $endgroup$
    – Henry
    Jan 4 at 15:16
















0












$begingroup$


Suppose I throw 3 fair dice 30 times.



Let,



X = no' of throws in which we don't get 4

Y = no' of throws in which we get 4 in only one die (out of 3)

Z = no' of throws in which we get 4 in exactly two dice (out of 3)



Compute $ Var(X+Y+Z) $



by definition $ Var(X+Y+Z) = E[(X+Y+Z)^2] + (E[X+Y+Z])^2 $



$ E[X+Y+Z] = E[X] + E[Y] + E[Z] = 29.8611$ because each variable of a multinomial distribution is of binomial distribution. so I know $ (E[X+Y+Z])^2 = 29.8611^2 $ but how do I calculate $ E[(X+Y+Z)^2] $ ?



Suggestions please?






probability variance expected-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Consider the probability of getting three $4$s and then consider $Var(n-X-Y-Z)$
    $endgroup$
    – Henry
    Jan 3 at 21:11








  • 1




    $begingroup$
    Can I just set new variable $ W = $ no' of throws in which I get up to two 4's and then calculate the variance of $ W $ as it is binomial variable?
    $endgroup$
    – bm1125
    Jan 4 at 12:49






  • 1




    $begingroup$
    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    $endgroup$
    – Henry
    Jan 4 at 15:16














0












0








0





$begingroup$


Suppose I throw 3 fair dice 30 times.



Let,



X = no' of throws in which we don't get 4

Y = no' of throws in which we get 4 in only one die (out of 3)

Z = no' of throws in which we get 4 in exactly two dice (out of 3)



Compute $ Var(X+Y+Z) $



by definition $ Var(X+Y+Z) = E[(X+Y+Z)^2] + (E[X+Y+Z])^2 $



$ E[X+Y+Z] = E[X] + E[Y] + E[Z] = 29.8611$ because each variable of a multinomial distribution is of binomial distribution. so I know $ (E[X+Y+Z])^2 = 29.8611^2 $ but how do I calculate $ E[(X+Y+Z)^2] $ ?



Suggestions please?






probability variance expected-value







share|cite|improve this question









$endgroup$




Suppose I throw 3 fair dice 30 times.



Let,



X = no' of throws in which we don't get 4

Y = no' of throws in which we get 4 in only one die (out of 3)

Z = no' of throws in which we get 4 in exactly two dice (out of 3)



Compute $ Var(X+Y+Z) $



by definition $ Var(X+Y+Z) = E[(X+Y+Z)^2] + (E[X+Y+Z])^2 $



$ E[X+Y+Z] = E[X] + E[Y] + E[Z] = 29.8611$ because each variable of a multinomial distribution is of binomial distribution. so I know $ (E[X+Y+Z])^2 = 29.8611^2 $ but how do I calculate $ E[(X+Y+Z)^2] $ ?



Suggestions please?






probability variance expected-value




probability variance expected-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 20:56









bm1125bm1125

65116




65116








  • 1




    $begingroup$
    Consider the probability of getting three $4$s and then consider $Var(n-X-Y-Z)$
    $endgroup$
    – Henry
    Jan 3 at 21:11








  • 1




    $begingroup$
    Can I just set new variable $ W = $ no' of throws in which I get up to two 4's and then calculate the variance of $ W $ as it is binomial variable?
    $endgroup$
    – bm1125
    Jan 4 at 12:49






  • 1




    $begingroup$
    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    $endgroup$
    – Henry
    Jan 4 at 15:16














  • 1




    $begingroup$
    Consider the probability of getting three $4$s and then consider $Var(n-X-Y-Z)$
    $endgroup$
    – Henry
    Jan 3 at 21:11








  • 1




    $begingroup$
    Can I just set new variable $ W = $ no' of throws in which I get up to two 4's and then calculate the variance of $ W $ as it is binomial variable?
    $endgroup$
    – bm1125
    Jan 4 at 12:49






  • 1




    $begingroup$
    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    $endgroup$
    – Henry
    Jan 4 at 15:16








1




1




$begingroup$
Consider the probability of getting three $4$s and then consider $Var(n-X-Y-Z)$
$endgroup$
– Henry
Jan 3 at 21:11






$begingroup$
Consider the probability of getting three $4$s and then consider $Var(n-X-Y-Z)$
$endgroup$
– Henry
Jan 3 at 21:11






1




1




$begingroup$
Can I just set new variable $ W = $ no' of throws in which I get up to two 4's and then calculate the variance of $ W $ as it is binomial variable?
$endgroup$
– bm1125
Jan 4 at 12:49




$begingroup$
Can I just set new variable $ W = $ no' of throws in which I get up to two 4's and then calculate the variance of $ W $ as it is binomial variable?
$endgroup$
– bm1125
Jan 4 at 12:49




1




1




$begingroup$
Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
$endgroup$
– Henry
Jan 4 at 15:16




$begingroup$
Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
$endgroup$
– Henry
Jan 4 at 15:16










2 Answers
2






active

oldest

votes


















1












$begingroup$

These $X$, $Y$, and $Z$ are strongly related to each other - to the point where we'd rather not work with them directly in calculating the variance. In particular, $X+Y+Z+W=30$, where $W$ is the number of throws in which all three dice come up fours. The variance of $X+Y+Z$ is the same as that of $W$. Can you calculate that?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
    $endgroup$
    – bm1125
    Jan 4 at 12:34










  • $begingroup$
    Another poster got there while I was sleeping in more detail, but it's a general rule; the variance of $aX+b$ is $|a|^2$ times the variance of $X$ for constants $a$ and $b$. We write $X+Y+Z=30-W$ and apply that.
    $endgroup$
    – jmerry
    Jan 4 at 19:50



















1












$begingroup$

Using the notation from @jmerry, let $W$ be the number of tosses in which all 3 dice show 4. Thus, $X+Y+Z+W=30$, i.e. $X+Y+Z=30-W$ implying $Var(X+Y+Z)=Var(30-W)$.



Let's consider $Q=30-W$.



$mathbb{E}[Q] = 30-mathbb{E}[W]$



$mathbb{E}[Q^2] = 900 + mathbb{E}[W^2] - 60mathbb{E}[W]$



$Var[Q] = mathbb{E}[Q^2] - mathbb{E}[Q]^2$



$=900 + mathbb{E}[W^2] - 60mathbb{E}[W] - (30-mathbb{E}[W])^2$



$=mathbb{E}[W^2] -mathbb{E}[W]^2$



$=Var(W)$ (by definition)



Thus, $Var(30-W) = Var(W) = Var(X+Y+Z)$.



$W$ follows a binomial distribution with parameters $N=30$ and $p = left( frac{1}{6} right)^3$. You can calculate the variance as $Np(1-p)$.






share|cite|improve this answer









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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

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    1












    $begingroup$

    These $X$, $Y$, and $Z$ are strongly related to each other - to the point where we'd rather not work with them directly in calculating the variance. In particular, $X+Y+Z+W=30$, where $W$ is the number of throws in which all three dice come up fours. The variance of $X+Y+Z$ is the same as that of $W$. Can you calculate that?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      $endgroup$
      – bm1125
      Jan 4 at 12:34










    • $begingroup$
      Another poster got there while I was sleeping in more detail, but it's a general rule; the variance of $aX+b$ is $|a|^2$ times the variance of $X$ for constants $a$ and $b$. We write $X+Y+Z=30-W$ and apply that.
      $endgroup$
      – jmerry
      Jan 4 at 19:50
















    1












    $begingroup$

    These $X$, $Y$, and $Z$ are strongly related to each other - to the point where we'd rather not work with them directly in calculating the variance. In particular, $X+Y+Z+W=30$, where $W$ is the number of throws in which all three dice come up fours. The variance of $X+Y+Z$ is the same as that of $W$. Can you calculate that?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      $endgroup$
      – bm1125
      Jan 4 at 12:34










    • $begingroup$
      Another poster got there while I was sleeping in more detail, but it's a general rule; the variance of $aX+b$ is $|a|^2$ times the variance of $X$ for constants $a$ and $b$. We write $X+Y+Z=30-W$ and apply that.
      $endgroup$
      – jmerry
      Jan 4 at 19:50














    1












    1








    1





    $begingroup$

    These $X$, $Y$, and $Z$ are strongly related to each other - to the point where we'd rather not work with them directly in calculating the variance. In particular, $X+Y+Z+W=30$, where $W$ is the number of throws in which all three dice come up fours. The variance of $X+Y+Z$ is the same as that of $W$. Can you calculate that?






    share|cite|improve this answer









    $endgroup$



    These $X$, $Y$, and $Z$ are strongly related to each other - to the point where we'd rather not work with them directly in calculating the variance. In particular, $X+Y+Z+W=30$, where $W$ is the number of throws in which all three dice come up fours. The variance of $X+Y+Z$ is the same as that of $W$. Can you calculate that?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 3 at 21:12









    jmerryjmerry

    13.6k1629




    13.6k1629












    • $begingroup$
      Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      $endgroup$
      – bm1125
      Jan 4 at 12:34










    • $begingroup$
      Another poster got there while I was sleeping in more detail, but it's a general rule; the variance of $aX+b$ is $|a|^2$ times the variance of $X$ for constants $a$ and $b$. We write $X+Y+Z=30-W$ and apply that.
      $endgroup$
      – jmerry
      Jan 4 at 19:50


















    • $begingroup$
      Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      $endgroup$
      – bm1125
      Jan 4 at 12:34










    • $begingroup$
      Another poster got there while I was sleeping in more detail, but it's a general rule; the variance of $aX+b$ is $|a|^2$ times the variance of $X$ for constants $a$ and $b$. We write $X+Y+Z=30-W$ and apply that.
      $endgroup$
      – jmerry
      Jan 4 at 19:50
















    $begingroup$
    Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
    $endgroup$
    – bm1125
    Jan 4 at 12:34




    $begingroup$
    Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
    $endgroup$
    – bm1125
    Jan 4 at 12:34












    $begingroup$
    Another poster got there while I was sleeping in more detail, but it's a general rule; the variance of $aX+b$ is $|a|^2$ times the variance of $X$ for constants $a$ and $b$. We write $X+Y+Z=30-W$ and apply that.
    $endgroup$
    – jmerry
    Jan 4 at 19:50




    $begingroup$
    Another poster got there while I was sleeping in more detail, but it's a general rule; the variance of $aX+b$ is $|a|^2$ times the variance of $X$ for constants $a$ and $b$. We write $X+Y+Z=30-W$ and apply that.
    $endgroup$
    – jmerry
    Jan 4 at 19:50











    1












    $begingroup$

    Using the notation from @jmerry, let $W$ be the number of tosses in which all 3 dice show 4. Thus, $X+Y+Z+W=30$, i.e. $X+Y+Z=30-W$ implying $Var(X+Y+Z)=Var(30-W)$.



    Let's consider $Q=30-W$.



    $mathbb{E}[Q] = 30-mathbb{E}[W]$



    $mathbb{E}[Q^2] = 900 + mathbb{E}[W^2] - 60mathbb{E}[W]$



    $Var[Q] = mathbb{E}[Q^2] - mathbb{E}[Q]^2$



    $=900 + mathbb{E}[W^2] - 60mathbb{E}[W] - (30-mathbb{E}[W])^2$



    $=mathbb{E}[W^2] -mathbb{E}[W]^2$



    $=Var(W)$ (by definition)



    Thus, $Var(30-W) = Var(W) = Var(X+Y+Z)$.



    $W$ follows a binomial distribution with parameters $N=30$ and $p = left( frac{1}{6} right)^3$. You can calculate the variance as $Np(1-p)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Using the notation from @jmerry, let $W$ be the number of tosses in which all 3 dice show 4. Thus, $X+Y+Z+W=30$, i.e. $X+Y+Z=30-W$ implying $Var(X+Y+Z)=Var(30-W)$.



      Let's consider $Q=30-W$.



      $mathbb{E}[Q] = 30-mathbb{E}[W]$



      $mathbb{E}[Q^2] = 900 + mathbb{E}[W^2] - 60mathbb{E}[W]$



      $Var[Q] = mathbb{E}[Q^2] - mathbb{E}[Q]^2$



      $=900 + mathbb{E}[W^2] - 60mathbb{E}[W] - (30-mathbb{E}[W])^2$



      $=mathbb{E}[W^2] -mathbb{E}[W]^2$



      $=Var(W)$ (by definition)



      Thus, $Var(30-W) = Var(W) = Var(X+Y+Z)$.



      $W$ follows a binomial distribution with parameters $N=30$ and $p = left( frac{1}{6} right)^3$. You can calculate the variance as $Np(1-p)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Using the notation from @jmerry, let $W$ be the number of tosses in which all 3 dice show 4. Thus, $X+Y+Z+W=30$, i.e. $X+Y+Z=30-W$ implying $Var(X+Y+Z)=Var(30-W)$.



        Let's consider $Q=30-W$.



        $mathbb{E}[Q] = 30-mathbb{E}[W]$



        $mathbb{E}[Q^2] = 900 + mathbb{E}[W^2] - 60mathbb{E}[W]$



        $Var[Q] = mathbb{E}[Q^2] - mathbb{E}[Q]^2$



        $=900 + mathbb{E}[W^2] - 60mathbb{E}[W] - (30-mathbb{E}[W])^2$



        $=mathbb{E}[W^2] -mathbb{E}[W]^2$



        $=Var(W)$ (by definition)



        Thus, $Var(30-W) = Var(W) = Var(X+Y+Z)$.



        $W$ follows a binomial distribution with parameters $N=30$ and $p = left( frac{1}{6} right)^3$. You can calculate the variance as $Np(1-p)$.






        share|cite|improve this answer









        $endgroup$



        Using the notation from @jmerry, let $W$ be the number of tosses in which all 3 dice show 4. Thus, $X+Y+Z+W=30$, i.e. $X+Y+Z=30-W$ implying $Var(X+Y+Z)=Var(30-W)$.



        Let's consider $Q=30-W$.



        $mathbb{E}[Q] = 30-mathbb{E}[W]$



        $mathbb{E}[Q^2] = 900 + mathbb{E}[W^2] - 60mathbb{E}[W]$



        $Var[Q] = mathbb{E}[Q^2] - mathbb{E}[Q]^2$



        $=900 + mathbb{E}[W^2] - 60mathbb{E}[W] - (30-mathbb{E}[W])^2$



        $=mathbb{E}[W^2] -mathbb{E}[W]^2$



        $=Var(W)$ (by definition)



        Thus, $Var(30-W) = Var(W) = Var(X+Y+Z)$.



        $W$ follows a binomial distribution with parameters $N=30$ and $p = left( frac{1}{6} right)^3$. You can calculate the variance as $Np(1-p)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 18:10









        Aditya DuaAditya Dua

        1,14418




        1,14418






























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