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The equations for getting the expected value of the binomial distribution in one of my textbooks is:

$$\sum_{x=i} x_i P(X=x_i)$$

and in another is:

$$\sum_x x P(x)$$

and both texts say that this actually equals:

$$\mu = np$$

and from what I can see the 'p' means the probability of success in the binomial equation.

As usual, I find the math notation of the summation equations is difficult to understand. If p = 0.6 and the number of trials is 4 then I from the above equation:

$$\mu = 4 \times 0.6 = 2.4$$

which seems simple, but when I look at the summation equations, I can't see that they are telling me to add up p 4 times because:

  1. there is a value $x_i$ to multiply a value $P(X = x_i)$
  2. What is this value $x_i$?
  3. Is $P(X = x_i)$ the same as 'p' ie 0.6 in the above example, or does it have a different value for each value of $x_i$ whatever that is?
  4. I tried summing the values of the discrete probability distribution and dividing by the number of different discrete values but didn't get the same answer.

I'm sure I'm not too far off getting this, but I don't see how the summation notation is saying the same thing as $\mu = np$.

Thanks in advance for any clarification.

Regards

Joe

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1 Answer 1

up vote 3 down vote accepted

If the number of trials is 4, then $X$ can take the values 0, 1, 2, 3 and 4. Then $x_1 = 0, x_2=1, \cdots, x_5 = 4$.

Since this is a binomial distribution, $P(X=x_i) = \binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}$, so $P(X=x_i)$ does not equal $p$ (at least not usually), and $P(X=x_i)$ has different values for different $x_i$.

Using the formula for finding $P(X=x_i)$ (provided above), you will be able to multiply $P(X=x_i)$ with $x_i$ for $i = 1, 2, 3, 4, 5$. When you add these five products together you will find the sum (at the top of your question) for the expectation value. If you do it right, I think you will find out that the sum is equal to $np$.

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OK brilliant. Thanks Eivind. That makes sense and I've got to the right answer now :) –  Joe Feb 16 '11 at 16:49
    
I'm glad I could help. –  please delete me Feb 16 '11 at 19:31

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