I think I know in general how to derive the maximum likelihood estimation for a parameter given a distribution. But I can't wrap my head around this one!

We have the observations: $(y_1, x_1),...,(y_n, x_n)$ from a random sample where: $Y_i \sim N(\beta x_i, 1)$ for $i = 1,2,...,n$. We are interested in the expected value of $y_i$ given $x_i$.

How do i derive the ML-estimate for the parameter $\beta?$

Don't know how to begin when the parameter $\beta$ shows up in that way.


I set:

$$\tilde{\mu} = x_i\beta$$

The Likelihood is:

$$\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}(\frac{y_i - \tilde{\mu}}{\sigma})^2} \iff (\frac{1}{\sqrt{2\pi\sigma^2}})^n e^{-\sum_{i=1}^{n}\frac{1}{2\sigma^2}(y_i - \tilde{\mu})^2}$$

Then the log-likelihood is:

$$-\frac{n}{2}\ln{2\pi} -\frac{n}{2}\ln{\sigma^2} -\frac{1}{2\sigma^2}\sum_{i=1}^{n}(y_i - \tilde{\mu})^2$$

If we take the derivative of the log-likelihood we get:

$$-\frac{1}{2\sigma^2}\frac{d}{d\tilde{\mu}}(\sum_{i=1}^{n}(y_i - \tilde{\mu})^2)$$

Which is:

$$-\frac{1}{2\sigma^2}\frac{d}{d\tilde{\mu}}(\sum_{i=1}^{n}(y_i^2 - 2y_i\tilde{\mu} + \tilde{\mu}^2))$$

Should i now insert $\tilde{\mu} = x_i\beta$?

If i do i get:

$$-\frac{1}{2\sigma^2}\frac{d}{d\beta}(\sum_{i=1}^{n}(y_i^2 - 2y_i x_i\beta + x_i^2\beta^2))$$

What should i do now, this approach below seems bad?

$$-\frac{1}{2\sigma^2}\sum_{i=1}^{n}\frac{d}{d\beta}(y_i^2 - 2y_i x_i\beta + x_i^2\beta^2)$$

$$-\frac{1}{2\sigma^2}\sum_{i=1}^{n} - 2y_i x_i + 2 x_i^2\beta$$

$$\frac{1}{2\sigma^2} 2n\bar{x} (n\bar{y} - n\bar{x}\beta)$$

And set to zero for finding the estimate: $$\frac{1}{2\sigma^2} 2n\bar{x} (n\bar{y} - n\bar{x}\beta) = 0$$

$$\bar{y} = \bar{x}\beta$$

$$\frac{\bar{y}}{\bar{x}} = \beta$$

Which is wrong.


I do try my best to understand so please explain if you downvote.

  • 2
    $\begingroup$ Start by writing out the likelihood. If you don't know how to do it for $\beta x_i$, write it as a product of $n$ factors each using $\mu_i$, then replace $\mu_i$ with $\beta x_i$ for all $i$. $\endgroup$
    – Sycorax
    Dec 1, 2022 at 19:01
  • 3
    $\begingroup$ 1. treat X as known, not a random variable. 2. Recall the chain rule. 3. It's always easiest to calculate MLEs from a single observation. $\endgroup$
    – AdamO
    Dec 1, 2022 at 19:07
  • 3
    $\begingroup$ In the expression for the normal distribution, you have $g(\beta)=(y_i - x_i \beta)^2$. What's the derivative of $g$ wrt $\beta$? $\endgroup$
    – Sycorax
    Dec 1, 2022 at 19:15
  • 2
    $\begingroup$ Nope. Intuitively, that doesn't make sense because it implies $\beta$ is not a real number when $\bar x$ is 0, and that the estimation would become very volatile for small changes in $\bar x$ around 0. Write out the steps from the start and don't skip any any of the algebra. $\endgroup$
    – Sycorax
    Dec 1, 2022 at 19:43
  • 2
    $\begingroup$ You go astray at the outset: $\tilde\mu$ depends on a single observation $x_i.$ That's not going to be useful and, as a notation, it's misleading you, because $\tilde\mu$ needs a subscript $i.$ You also introduce an undefined and superfluous quantity "$\sigma,$" further confusing the issue. Since the log likelihood of $y_i$ is $-(y_i-\beta x_i)^2/2$ (up to an additive constant), use that to form the log likelihood for an independent set of $(x_i,y_i)$ data and go on from there. The rest can be done with elementary algebra -- you don't even need Calculus. $\endgroup$
    – whuber
    Dec 1, 2022 at 20:39

1 Answer 1


The derivation is basically fine until here. After this step, you introduce $\bar x, \bar y$ and the result is not correct. $$0=-\frac{1}{2\sigma^2}\sum_{i=1}^{n} - 2y_i x_i + 2 x_i^2\beta.$$ It's given that $\sigma=1$. We can multiply through by the leading fraction. The rest is rearranging. $$\begin{align} 0 &= \sum_{i=1}^{n} y_i x_i - x_i^2\beta\\ &= \sum_{i=1}^{n} y_i x_i - \sum_{j=1}^n x_j^2\beta \\ \sum_{i=1}^{n} y_i x_i &= \beta \sum_{j=1}^n x_j^2 \\ \frac{\sum_{i=1}^{n} y_i x_i}{ \sum_{j=1}^n x_j^2} &= \beta \end{align}$$


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