I'm currently assigning a introductory stats class, and I just can't seem to find out when to use the different variance identities. I have provided an example of an assignment where I got it wrong, but I just can't figure out why its wrong.


Given two random variables: X ~ N(20,5^2) and Y ~ N(50,10^2). The random variables X and Y describe the weigh of rocks loaded from two different piles, to a truck. (X , Y is i.d.d). Okay, the assignment says, given that you load 10 rocks from each pile (X, Y), what is the probability that the weigh exceeds 800kg?

New random variable: Z = 10*X + 10*Y.

Calculate the estimate mean: E(Z) = E(10X) + E(10Y) = 10*20 + 50*10 = 700.

Now, heres where I get it wrong:

Calculate the variance: Var(Z)= Var(10*X+10*Y) = 10^2*Var(X) + 10^2*Var(Y) = 12500

Then Z ~ N(700,12500).

My teacher provided a result, and the result is to sum up the variances, heck the result is: Var(Z) = sum(1..10, Var(X)) + sum(1..10, Var(Y)) = 10*Var(X) + 10*Var(Y) = 1250.

I just can't seem to figure when to sum up the variances, and when to use the variance identity as my teacher call it (square the constant in front of the variance).

  • $\begingroup$ The hyphen is used to indicate subtraction. If you're indicating the distribution of a variable, you should use the tilda ~ $\endgroup$ – Acccumulation May 20 at 14:54

I don't know if this is a lack of rigor in your teacher's wording, or you copied the problem incorrectly, but if it is indeed true that "The random variables X and Y describe the weigh[t] of rocks loaded from two different piles", then your calculation is correct. However, to say "The random variables X and Y describe the weigh[t] of rocks loaded from two different piles" is to say that each pile has a particular weight that is distributed according to the respective distribution; that is, for each pile, one number is chosen according to the distribution, and each rock in that pile is then equal to that one number.

There is an important difference between "A, B = X ~ N" and "A, B ~ N". The first says that A and B are both equal to the same number drawn from a normal distribution, while the second says that A and B are different numbers both drawn from the same normal distribution. From context, it is reasonable to consider the rocks to be the second case rather than the first, in which case the variance should be multiplied by 10. Even when two random variables have the exact same distribution, they need to be treated as different variables, so in cases like this we calculate the variance correctly.

You further say "(X , Y [are] i.d.d)". But you gave different distributions for X and Y, so X and Y are not identical to each other. Presumably, this means that there is a collection of $X_i$ that are i.d.d. So a more rigorous way of saying this is:

Given two sets of random variables: $X_i$ ~ $N(20,5^2)$ and $Y_i$ ~ $N(50,10^2)$ that describe the weight of rocks loaded from two different piles to a truck, where $X_i$ and $Y_i$ are i.d.d., if you load 10 rocks from each pile (X, Y), what is the probability that the weight exceeds 800kg?

So here, you have ten copies of $N(20,5^2)$, not one $N(20,5^2)$ that has been multiplied by 10. If you had two rocks, you would have $N(20,5^2)+N(20,5^2)$, and that is not equal to $N(20+20,5^2+5^2)$. Similarly, ten copies of $N(20,5^2)$ is not the same as $N(20*10,5^2*10)$.

So, to answer your question: if things are phrased rigorously, then you can use the rule $var(10X)= 100 var(X)$. If things are not phrased rigorously, you have to think about whether $10X$ refers to one instance of a random variable that has been multiplied by $10$ (in which case you can use the rule $var(10X)= 100 var(X)$), or whether it refers to ten different instances of random variables, each drawn separately (in which case $var(10X)=10var(X)$). In this case, the fact that the variables were stated as being i.d.d. makes it clear that they are separate instances; $X$ and $Y$ can't be i.d.d. with respect to each other, and it makes no sense to say that a single instance of a variable is i.d.d., so they must be multiple instances from the same distribution.

You should look carefully at how the problem was phrased, and possibly have a discussion with your instructor about this distinction. Either your instructor didn't phrase this rigorously, or they didn't discuss this distinction, or they did discuss it but it didn't stick with you. Either way, they should get feedback about your confusion.

  • $\begingroup$ That was very helpful, thanks a lot! I think the problem occurs due to, that we spent next to zero time about this distinction, as well as they never taught us to read statistic notation nor how/when to use different identities for the mean, variance etc. The course is mainly focusing on hypothesis testing, anova, linear regression - like how to use statistics. $\endgroup$ – Oliver Bak May 22 at 18:24

There is a difference, between a sum of random variables, and multiplying a constant to a random variable.

In this case, each rock's weight is a random variable so the problem is a sum of random variables. In this case, is:

Let $X_1 ... X_{10}, Y_1 ... Y_{10}$ be random variables. The random variable you are looking for is Z = $\Sigma^{10}_{i = 1} X_i + \Sigma^{10}_{i = 1} Y_i$. Which variance is the same as your teacher's approach since they are independent.


In addition to what @Kane_Chua says, i.e. $10X\neq \sum_{i=1}^{10} X_i$; your result has an intuitive explanation. If you choose a rock randomly and multiply it with $10$, the variance you get will be larger than choosing $10$ different rocks and summing them. In the former, your calculation depends on one rock; where in the latter, it depends on $10$ rocks, which is the key factor that reduces your variance ten times.


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