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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?

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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 – 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
up vote 2 down vote accepted

The mean of $x$ = mean of $y$

This is not true.

The way you should approach this problem is to use the formulas for mean and standard deviation directly: \begin{align*} \text{Mean}(y_1, y_2, \ldots, y_n) &= \frac{y_1 + y_2 + \cdots + y_n}{n} \\ &= \frac{(ax_1 + b) + (ax_2 + b) + \cdots + (ax_n + b)}{n} \\ &= \frac{a(x_1 + x_2 + \cdots + x_n) + nb}{n} \\ &= a \cdot \text{Mean}(x_1, x_2, \ldots, x_n) + b \\ \end{align*}

See if you can do a similar algebraic manipulation for standard deviation.

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nice answer @goos – Sameh Shenawy Feb 6 '14 at 23:52
Wow, this actually makes a lot of sense all of a sudden. So for standard deviation, I'd pretty much be using the formula for standard deviation and finding a way to correlate x and y? I think I understand this question now – Valrok Feb 6 '14 at 23:56

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

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