1
$\begingroup$

This is the question I'm answering:

$$ \text{Suppose that the method of maximum likelihood is used to fit the following model representing exponential growth,}\; Y_i = \beta_1 e^{x_{i}} + \epsilon_i \; \text{to the data} \; (x_i, y_i), \; \text{where} \; i = 1,...,n \; \text{and} \; \epsilon_i \sim^{inep}N(0, \sigma^2). \; \text{It follows that} \; Y_i \sim^{inep}N(\beta_1e^{x_{i}}, \sigma^2). \; \text{The maximum likelihood estimator of}\; \beta_1 \; \text{is}\\ \hat{\beta_1} = \frac{\sum_{i=1}^n Y_ie^{x_i}}{\sum_{i=1}^n e^{2x_i}}\\ \text{Derive} \; E[\hat{\beta_1}] \; \text{and}\; Var(\hat{\beta_1}) \; \text{and hence write down the distribution of}\; \hat{\beta_1} $$ So, the expectation was easy enough to derive, $E[\hat{\beta_1}] = \beta_1$. However, in deriving the variance I'm left with a sum in the denominator that I can't seem to simplify? $$\begin{align} Var(\hat{\beta_1}) &= Var\left(\frac{\sum_{i=1}^n Y_ie^{x_i}}{\sum_{i=1}^n e^{2x_i}}\right)\\ &= \frac{\sum_{i=1}^n e^{2x_i}}{\sum_{i=1}^n e^{4x_i}}Var(Y_i)\\ &= \frac{\sigma^2}{\sum_{i=1}^n e^{2x_i}} \end{align}$$ Is there any way to simplify more? I'm also not sure how to determine the distribution. Can I just treat the $e^{x_i}$ terms as constant and thus $\hat{\beta_1} \sim N\left(\beta_1, \frac{\sigma^2}{\sum_{i=1}^n e^{2x_i}}\right)$ ?

$\endgroup$
2
  • $\begingroup$ Note $(\sum e^{2x_i})^2 \neq \sum e^{4x_i}$ $\endgroup$
    – Golden_Ratio
    Mar 7 at 15:33
  • $\begingroup$ How is it different from properties of the same estimate for linear model $Y_i = \beta_1 v_i + \epsilon_i$, where $v_i = e^{x_i}$? $\endgroup$
    – Yalikesifulei
    Mar 7 at 17:01

1 Answer 1

Reset to default
1
$\begingroup$

Variance: \begin{align} var\left(\frac{\sum Y_i e^{x_i}}{\sum e^{2x_i}}\right) = \frac{1}{ (\sum e^{2x_i})^2} var\left(\sum Y_i e^{x_i}\right) = \frac{1}{ (\sum e^{2x_i})^2} \sum e^{2x_i} var(Y_i) = \frac{ \sigma^2 }{ \sum e^{2x_i}} \end{align}

Distribution: \begin{align} Y_i &\sim N(\beta e^{x_i}, \sigma^2)\\ e^{x_i}Y_i &\sim N(\beta e^{2x_i}, e^{2x_i}\sigma^2)\\ \sum e^{x_i}Y_i &\sim N(\beta \sum e^{2x_i}, \sum e^{2x_i}\sigma^2)\\ \frac{\sum e^{x_i}Y_i}{\sum e^{2x_i}} &\sim N(\beta \sum e^{2x_i}/\sum e^{2x_i}, \sum e^{2x_i}\sigma^2/\sum e^{2x_i})\\ \hat \beta & \sim N(\beta, \sigma ^ 2/ \sum e^{2x_i}) \end{align} that is basically just a linear combination of a normal random variables $Y_i$, $i=1,...,n$.

$\endgroup$
4
  • 1
    $\begingroup$ In your second step, you assert that $\left(\sum e^{x_i}\right)^2 = \sum e^{2x_i}$ by pulling that term outside of the variance. Is this accurate? I'm just asking because this conflicts with the first comment left on my question so I'm a bit confused $\endgroup$
    – spooleey
    Mar 9 at 12:04
  • $\begingroup$ @spooleey, which step do you refer to? But, anyway, the answer is no $var(\sum Y_i e^{xi}) = \sum e^{2x_i} var(Y_i) = Var(Y) \sum e^{2x_i}$. $\endgroup$
    – V. Vancak
    Mar 9 at 13:03
  • $\begingroup$ I mean exactly that step. When you pull a constant out of a variance, you square it, so $Var(\sum Y_i e^{x_i}) = \sum e^{2xi} Var(Y_i) \implies (\sum e^{x_i})^2 = \sum e^{2x_i}$ no? Otherwise, where is $\sum e^{2x_i}$ coming from? @V.Vancak $\endgroup$
    – spooleey
    Mar 11 at 9:52
  • 1
    $\begingroup$ @spooleey No. for independent $X$ and $Y$, $var(aX + bY) = a ^2 var(X) + b^2 var(Y)$. In our case, $var(Y_1e^{x_1} + Y_2e^{x_2}) = (e^{x_1})^2var(Y_1) + (e^{x_2})^2var(Y_2) = e^{2x_1}var(Y_1) + e^{2x_2})var(Y_1)$. Namely, the property holds for every summand separately but not for the whole sum. $\endgroup$
    – V. Vancak
    Mar 11 at 10:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .