# 26.1 - Sums of Independent Normal Random Variables

26.1 - Sums of Independent Normal Random VariablesWell, we know that one of our goals for this lesson is to find the probability distribution of the sample mean when a random sample is taken from a population whose measurements are normally distributed. Then, let's just get right to the punch line! Well, first we'll work on the probability distribution of a linear combination of independent normal random variables \(X_1, X_2, \ldots, X_n\). On the next page, we'll tackle the sample mean!

If \(X_1, X_2, \ldots, X_n\) >are mutually independent normal random variables with means \(\mu_1, \mu_2, \ldots, \mu_n\) and variances \(\sigma^2_1,\sigma^2_2,\cdots,\sigma^2_n\), then the linear combination:

\(Y=\sum\limits_{i=1}^n c_iX_i\)

follows the normal distribution:

\(N\left(\sum\limits_{i=1}^n c_i \mu_i,\sum\limits_{i=1}^n c^2_i \sigma^2_i\right)\)

### Proof

We'll use the moment-generating function technique to find the distribution of \(Y\). In the previous lesson, we learned that the moment-generating function of a linear combination of independent random variables \(X_1, X_2, \ldots, X_n\) >is:

\(M_Y(t)=\prod\limits_{i=1}^n M_{X_i}(c_it)\)

Now, recall that if \(X_i\sim N(\mu, \sigma^2)\), then the moment-generating function of \(X_i\) is:

\(M_{X_i}(t)=\text{exp} \left(\mu t+\dfrac{\sigma^2t^2}{2}\right)\)

Therefore, the moment-generating function of \(Y\) is:

\(M_Y(t)=\prod\limits_{i=1}^n M_{X_i}(c_it)=\prod\limits_{i=1}^n \text{exp} \left[\mu_i(c_it)+\dfrac{\sigma^2_i(c_it)^2}{2}\right] \)

Evaluating the product at each index \(i\) from 1 to \(n\), and using what we know about exponents, we get:

\(M_Y(t)=\text{exp}(\mu_1c_1t) \cdot \text{exp}(\mu_2c_2t) \cdots \text{exp}(\mu_nc_nt) \cdot \text{exp}\left(\dfrac{\sigma^2_1c^2_1t^2}{2}\right) \cdot \text{exp}\left(\dfrac{\sigma^2_2c^2_2t^2}{2}\right) \cdots \text{exp}\left(\dfrac{\sigma^2_nc^2_nt^2}{2}\right) \)

Again, using what we know about exponents, and rewriting what we have using summation notation, we get:

\(M_Y(t)=\text{exp}\left[t\left(\sum\limits_{i=1}^n c_i \mu_i\right)+\dfrac{t^2}{2}\left(\sum\limits_{i=1}^n c^2_i \sigma^2_i\right)\right]\)

Ahaaa! We have just shown that the moment-generating function of \(Y\) is the same as the moment-generating function of a normal random variable with mean:

\(\sum\limits_{i=1}^n c_i \mu_i\)

and variance:

\(\sum\limits_{i=1}^n c^2_i \sigma^2_i\)

Therefore, by the uniqueness property of moment-generating functions, \(Y\) must be normally distributed with the said mean and said variance. Our proof is complete.

## Example 26-1

Let \(X_1\) be a normal random variable with mean 2 and variance 3, and let \(X_2\) be a normal random variable with mean 1 and variance 4. Assume that \(X_1\) and \(X_2\) are independent. What is the distribution of the linear combination \(Y=2X_1+3X_2\)?

#### Solution

The previous theorem tells us that \(Y\) is normally distributed with mean 7 and variance 48 as the following calculation illustrates:

\((2X_1+3X_2)\sim N(2(2)+3(1),2^2(3)+3^2(4))=N(7,48)\)

What is the distribution of the linear combination \(Y=X_1-X_2\)?

#### Solution

The previous theorem tells us that \(Y\) is normally distributed with mean 1 and variance 7 as the following calculation illustrates:

\((X_1-X_2)\sim N(2-1,(1)^2(3)+(-1)^2(4))=N(1,7)\)

## Example 26-2

History suggests that scores on the Math portion of the Standard Achievement Test (SAT) are normally distributed with a mean of 529 and a variance of 5732. History also suggests that scores on the Verbal portion of the SAT are normally distributed with a mean of 474 and a variance of 6368. Select two students at random. Let \(X\) denote the first student's Math score, and let \(Y\) denote the second student's Verbal score. What is \(P(X>Y)\)?

#### Solution

We can find the requested probability by noting that \(P(X>Y)=P(X-Y>0)\), and then taking advantage of what we know about the distribution of \(X-Y\). That is, \(X-Y\) is normally distributed with a mean of 55 and variance of 12100 as the following calculation illustrates:

\((X-Y)\sim N(529-474,(1)^2(5732)+(-1)^2(6368))=N(55,12100)\)

Then, finding the probability that \(X\) is greater than \(Y\) reduces to a normal probability calculation:

\begin{align} P(X>Y) &=P(X-Y>0)\\ &= P\left(Z>\dfrac{0-55}{\sqrt{12100}}\right)\\ &= P\left(Z>-\dfrac{1}{2}\right)=P\left(Z<\dfrac{1}{2}\right)=0.6915\\ \end{align}

That is, the probability that the first student's Math score is greater than the second student's Verbal score is 0.6915.

## Example 26-3

Let \(X_i\) denote the weight of a randomly selected prepackaged one-pound bag of carrots. Of course, one-pound bags of carrots won't weigh exactly one pound. In fact, history suggests that \(X_i\) is normally distributed with a mean of 1.18 pounds and a standard deviation of 0.07 pound.

Now, let \(W\) denote the weight of randomly selected prepackaged three-pound bag of carrots. Three-pound bags of carrots won't weigh exactly three pounds either. In fact, history suggests that \(W\) is normally distributed with a mean of 3.22 pounds and a standard deviation of 0.09 pound.

Selecting bags at random, what is the probability that the sum of three one-pound bags exceeds the weight of one three-pound bag?

#### Solution

Because the bags are selected at random, we can assume that \(X_1, X_2, X_3\) and \(W\) are mutually independent. The theorem helps us determine the distribution of \(Y\), the sum of three one-pound bags:

\(Y=(X_1+X_2+X_3) \sim N(1.18+1.18+1.18, 0.07^2+0.07^2+0.07^2)=N(3.54,0.0147)\)

That is, \(Y\) is normally distributed with a mean of 3.54 pounds and a variance of 0.0147. Now, \(Y-W\), the difference in the weight of three one-pound bags and one three-pound bag is normally distributed with a mean of 0.32 and a variance of 0.0228, as the following calculation suggests:

\((Y-W) \sim N(3.54-3.22,(1)^2(0.0147)+(-1)^2(0.09^2))=N(0.32,0.0228)\)

Therefore, finding the probability that \(Y\) is greater than \(W\) reduces to a normal probability calculation:

\begin{align} P(Y>W) &=P(Y-W>0)\\ &= P\left(Z>\dfrac{0-0.32}{\sqrt{0.0228}}\right)\\ &= P(Z>-2.12)=P(Z<2.12)=0.9830\\ \end{align}

That is, the probability that the sum of three one-pound bags exceeds the weight of one three-pound bag is 0.9830. Hey, if you want more bang for your buck, it looks like you should buy multiple one-pound bags of carrots, as opposed to one three-pound bag!