I am trying to figure out why calculating the SE directly in 22-4 gives me a larger result than that given in the solution (examples change slightly each time the page loads): http://www.stat.berkeley.edu/~stark/SticiGui/Text/standardError.htm#SE_of_RV

In 22-3, the method for directly calculating the SE seems to be $SE(X) = \sqrt{\sum \limits_{i=1}^{n}P(X=x_i)(x_i-E(X))^2} $, where you multiply each specific outcome by its probability, sum those values and then take the square root.

The specific values I had for 22-4 were:

You pay $\$2$ to roll a fair pair of dice. You win $\$3$ if you roll doubles. The SE of your net winnings after nine plays of the game is __?

In 22-4, the solution calculates the SE of the binomial and then uses a linear transformation to get the standard error for playing with dollar amounts other than 0 and 1:

$Y= -18+3X$ the linear relationship of net winnings for 9 plays, where X is the number of wins in 9 plays.

$SE(X)=\sqrt{np(1-p)}=\sqrt{9\frac{1}{6}\frac{5}{6}}= 1.118$ The SE of a binomial distribution for 9 plays.

$SE(Y)=|3|SE(X)=\color{blue}{3.354}$ The SE for 9 plays if you win $\$3$ by rolling doubles and lose $\$2$ otherwise.

I tried to use the method I used for 22-3 (multiply the squared deviation of each outcome by its probability, sum those values and take the square root to get the SE), however I'm not sure why this doesn't work.

$E(X) = x_0P(X=x_0)+x_1P(X=x_1)= -2(5/6)+3(1/6)=-7/6$

$SE(X) = \sqrt{P(X=x_0)(x_0 - E(X))^2+P(X=x_1)(x_1 - E(X))^2}$

$=\sqrt{\frac{5}{6}(-2+\frac{7}{6})^2+\frac{1}{6}(3+\frac{7}{6})^2}= 1.863$

for nine plays $\sqrt{9}SE(X)=\color{red}{5.5902}$

  • 1
    $\begingroup$ After you have paid two dollars and won three dollars, your net winnings are one dollar, not three. $\endgroup$ – whuber May 8 '13 at 20:46
  • $\begingroup$ You have defined $X$ to be the number of wins in 9 plays. The realizations of $X$ are then in the set {0,1,2,3,4,5,6,7,8,9}. $\endgroup$ – soakley May 8 '13 at 21:14

You need to be doing your direct method on $Y$. $Y$ takes on values in the set $\{-18, -15, -12, -9, -6, -3, 0, 3, 6, 9\}$ and if you work it out, you will find $E[Y] = -13.5$. Using the binomial probabilities and your equation, you will find $\sigma_Y^2 = 45/4$ , or $\sigma_Y = 3.354$.

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