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












0














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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  • 1




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








  • 1




    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?
    – bm1125
    2 days ago






  • 1




    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    – Henry
    2 days ago
















0














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


















  • 1




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








  • 1




    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?
    – bm1125
    2 days ago






  • 1




    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    – Henry
    2 days ago














0












0








0







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













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













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share|cite|improve this question




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asked Jan 3 at 20:56









bm1125

63516




63516








  • 1




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








  • 1




    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?
    – bm1125
    2 days ago






  • 1




    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    – Henry
    2 days ago














  • 1




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








  • 1




    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?
    – bm1125
    2 days ago






  • 1




    Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
    – Henry
    2 days ago








1




1




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






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






1




1




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?
– bm1125
2 days ago




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?
– bm1125
2 days ago




1




1




Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
– Henry
2 days ago




Yes - that works very well since each set of throws is independent and gives a yes/no answer to "up to two 4s"
– Henry
2 days ago










2 Answers
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oldest

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1














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





















  • Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
    – bm1125
    2 days ago










  • 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.
    – jmerry
    2 days ago



















1














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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    2 Answers
    2






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    active

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    active

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    1














    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





















    • Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      – bm1125
      2 days ago










    • 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.
      – jmerry
      2 days ago
















    1














    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





















    • Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      – bm1125
      2 days ago










    • 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.
      – jmerry
      2 days ago














    1












    1








    1






    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












    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









    jmerry

    2,276210




    2,276210












    • Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      – bm1125
      2 days ago










    • 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.
      – jmerry
      2 days ago


















    • Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
      – bm1125
      2 days ago










    • 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.
      – jmerry
      2 days ago
















    Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
    – bm1125
    2 days ago




    Thanks for the answer. But why is the variance of $ X+Y+Z $ be the same as that of $ W $ ?
    – bm1125
    2 days ago












    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.
    – jmerry
    2 days ago




    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.
    – jmerry
    2 days ago











    1














    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


























      1














      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
























        1












        1








        1






        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












        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 2 days ago









        Aditya Dua

        86418




        86418






























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