# Sums of Independent Normal Random Variables

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Well, 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 X1X2, ... , Xn. On the next page, we'll tackle the sample mean!

 Theorem: If X1, X2, ... , Xn are mutually independent normal random variables with means μ1, μ2, ... , μn 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 X1X2, ... , Xn is:

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

Now, recall that if Xi ~ N(μ, σ2), then the moment-generating function of Xi 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

Let X1 be a normal random variable with mean 2 and variance 3, and let X2 be a normal random variable with mean 1 and variance 4. Assume that Xand X2 are independent. What is the distribution of the linear combination Y = 2X+ 3X2?

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 = X1 − X2?

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

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 (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

Let Xi 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 Xi 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 X1, X2, X3, 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, YW, 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 is greater than 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!