# A Differential of Multiple Linear Regression

1. Oct 4, 2017

### fny

Say you have a log-level regression as follows:

$$\log Y = \beta_0 + \beta_1 X_1 + \beta_1 X_2 + \ldots + \beta_n X_n$$

We're trying come up with a meaningful interpretation for changes Y due to a change in some Xk.

If we take the partial derivative with respect to Xk. we end up with

$$\frac{dY}{Y} = \beta_k \cdot dX_k$$

which implies that if Xk. increases by 1, you expect Y to increase by 100βk percent.

Can someone walk through the calculus to get from this

$$\frac{\partial}{\partial X_k} \log{y}= \frac{\partial}{\partial X_k} (\beta_0 + \beta_1 X_1 + \beta_1 X_2 + \ldots + \beta_n X_n)$$

to this

$$\frac{dY}{Y} = \beta_k dX_k$$?

I'm particularly confused about how one transitions from a partial derivate to a total derivative.

2. Oct 4, 2017

### andrewkirk

In general one cannot make that transition. The second last formula is correct but the last is not, unless there is no dependence between $X_k$ and any of the other $X_j$s. A corrected version of the last formula is:
$$\frac{dY}{Y} = \sum_{j=1}^n \beta_j dX_j$$

To get the total derivative wrt $X_k$ we use the total derivative formula, for the case where $Y$ is a function of $X_1,...,X_n$:

$$\frac{dY}{dX_k}=\sum_{j=1}^n \frac{\partial Y}{\partial X_j} \frac{dX_j}{dX_k}$$

In this case we have $Y = \exp\left(\beta_0 + \sum_{k=1}^j \beta_j X_j\right)$ so that $\frac{\partial Y}{\partial X_j} = \beta_jY$, and we also have $\frac{d X_k}{d X_k}=1$, so that the total derivative becomes:
$$\frac{dY}{dX_k}=Y\left(\beta_k + \sum_{\substack{j=1\\j\neq k}}^n \beta_j \frac{dX_j}{dX_k}\right)$$

This reduces to the formula you wrote above for the total derivative if all the $\frac{dX_j}{dX_k}$ are zero, ie if there are no dependences between $X_k$ and any of the other $X_j$.

What we can say is that $Y$ increases by $Y\beta_k\delta X_k$ if $X_k$ increases by $\delta X_k$ and all other variables do not change.