# Relationships between mean and standard deviation when one variable is linear function of another

Let $a$ and $b$ be constants and let $y_j = ax_j + b$ for $j=1,2,\ldots,n$.

What are the relationships between the means of $ya$ and $x$, and the standard deviations of $y$ and $x$?

I'm slightly confused with how to approach a theoretical question such as this and was wondering if anyone could help provide me some advice on how to approach this problem.

At the moment here is what I'm thinking, but I'm currently working without certainty:

We know

• $x_j = (y_j - b)/a$

• The mean of $x$ = mean of $y$

In terms of standard deviation, I'm not sure how they correlate at all right now aside from the fact that you need the mean of $x$ or $y$ in order to calculate the corresponding standard deviation.

If someone could help explain this question and help me understand what I'm being asked and how to solve this I would greatly appreciate it!

EDIT: So looking at the second portion of the question I am doing the following:

SD = sqrt(Sigma(y_i - y)^2/(n-1)))

SD(y) = (Sigma(yi - (ax+b)))/(n-1)

SD(y) = (Sigma (ax+b) - (ax+b))/(n-1)

SD(y) = 1/(n-1)

Is the following correct?

• Do you mean $ya$ or just $y$ in your second line? You also use $x$ and $y$ in many places but do you mean $x_j$ and $y_j$? – TooTone Feb 6 '14 at 23:51
• When $y=f(a,b,c)$, we have $\bar y=f(\bar a,\bar b,\bar c)$ (for any number of variables). The error $\sigma_y$ can be found using propagation of uncertainties:en.wikipedia.org/wiki/Propagation_of_uncertainty – Ragnar Feb 6 '14 at 23:51
• It's generally true that $\text{E}[aX+b]= a\text{E}[X]+b$ and that $\text{Var}[aX+b]=a^2\text{Var}[X]$. The latter means, in your notation, that $\text{SD}[aX+b]=a^2\text{SD}[X]$. – MPW Feb 7 '14 at 0:36
• That should be $a\text{SD}[X]$ at the end, couldn't edit the comment for some reason. – MPW Feb 7 '14 at 0:44

The mean of $x$ = mean of $y$
Since $y_j = a\cdot x_j + b$, then $$\bar{y}=\frac{\Sigma ( a\cdot x_j + b)}{n} \\ =\frac{a\Sigma \cdot x_j + nb}{n} \\=a\bar{x}+b$$ Use the same way with SD