I was reading Andrew Ng's CS229 lecture notes (page 12) about justifying squared loss risk as a means of estimating regressions parameters.

Andres explains that we first need to assume that the target function $y^{(i)}$ can be written as:

$$ y^{(i)} = \theta^Tx^{(i)} + \epsilon^{(i)}$$

where $e^{(i)}$ is the error term that captures unmodeled effects and random noise. Further assume that this noise is distributed as $\epsilon^{(i)} \sim \mathcal{N}(0, \sigma^2)$. Thus:

$$p(e^{(i)}) = \frac{1}{\sqrt{2\pi\sigma}}\exp \left( \frac{-(e^{(i)})^2}{2\sigma^2} \right) $$

Thus we can see that the error term is a function of $y^{(i)}$, $x^{(i)}$ and $\theta$ as in:

$$e^{(i)} = f(y^{(i)}, x^{(i)}; \theta) = y^{(i)} - \theta^Tx^{(i)}$$

thus we can substitute to the above equation for $e^{(i)}$

$$p(y^{(i)} - \theta^Tx^{(i)}) = \frac{1}{\sqrt{2\pi\sigma}}\exp \left( \frac{-(y^{(i)} - \theta^Tx^{(i)})^2}{2\sigma^2} \right)$$

Now we know that:

$p(e^{(i)}) = p(y^{(i)} - \theta^Tx^{(i)}) = p(f(y^{(i)}, x^{(i)}; \theta))$

Which is a function of the random variables $x^{(i)}$ and $y^{(i)}$ (and the non random variable $\theta$). Andrew then favors $x^{(i)}$ as being the conditioning variable and says:

$p(e^{(i)}) = p(y^{(i)} \mid x^{(i)})$

However, I can't seem to justify why we would favor expressing $p(e^{(i)})$ as $p(y^{(i)} \mid x^{(i)})$ and not the other way round $p(x^{(i)} \mid y^{(i)})$.

The problem I have his derivation is that with only the distribution for the error (which for me, seems to be symmetric wrt to x and y):

$$\frac{1}{\sqrt{2\pi\sigma}}\exp \left( \frac{-(e^{(i)})^2}{2\sigma^2} \right)$$

I can't see why we would favor $p(e^{(i)})$ as $p(y^{(i)} \mid x^{(i)})$ and not the other way round $p(x^{(i)} \mid y^{(i)})$ (just because we are interested in y, is not enough for me as a justification because just because that is our quantity of interest, it does not mean that the equation should be the way we want it to, i.e. it doesn't mean that it should be $p(y^{(i)} \mid x^{(i)})$, at least that doesn't seem to be the case from a purely mathematical perspective for me).

Another way of expressing my problem is the following:

The Normal equation seems to be symmetrical in $x^{(i)}$ and $y^{(i)}$. Why favor $p(y^{(i)} \mid x^{(i)})$ and not $p(x^{(i)} \mid y^{(i)})$. Furthermore, if its a supervised learning situation, we would get both pairs $(x^{(i)}, y^{(i)})$, right? Its not like we get one first and then the other.

Basically, I am just trying to understand why $p(y^{(i)} \mid x^{(i)})$ is correct and why $p(x^{(i)} \mid y^{(i)})$ is not the correct substitution for $p(e^{(i)})$.

  • 3
    $\begingroup$ Because you're trying to predict $y$ from $x$? $\endgroup$
    – Glen_b
    Aug 15, 2014 at 3:03
  • $\begingroup$ That might be our goal, but that doesn't justify $p(e^{(i)})$ to be what we want it to be. How do I know its not $p(x^{(i)}|y^{(i)} ; \theta)$ and I don't need some extra mathematical steps to get what I am actually interested in...? $\endgroup$ Aug 15, 2014 at 3:05
  • 3
    $\begingroup$ I really don't get your question at all. Predicting y given x is the starting point, and that tells you you're trying to evaluate $p(y|x)$, which then has the form you describe. $\endgroup$
    – Glen_b
    Aug 15, 2014 at 3:19
  • 1
    $\begingroup$ You're assuming there's random observation error in the $x$? That would be a different model, sometimes called errors-in-variables among other names. $\endgroup$
    – Glen_b
    Sep 7, 2016 at 22:06
  • 1
    $\begingroup$ The regression model conditions on $x$. If you don't want to do that, you don't use a model that does it. However even if you're trying to model a joint distribution, a conditional distribution can be relevant since you can write P(x,y) = P(y|x) P(x) $\endgroup$
    – Glen_b
    Sep 8, 2016 at 15:53

7 Answers 7


Overall, you're correct; $p(x|y)$ will be a normally-distributed function of the size of the error. However, in general, you will be using multiple exogenously fixed input variables $x$ to predict a single output variable $y$, so we're rarely interested in guessing $x$ directly based on what we know about $y$.

An example will be helpful here: Suppose you have a set of pictures of animals and you want to know the type of animal present in each picture. Your $x$ will be an image, and $y$ will be the type of animal in the image. $p(y|x)$ makes a lot of sense--we're trying to find probabilistically the correct class label for each image.

$p(x|y)$ is kind of odd. It's a probability of a single image, given that the image's label is a cat. If you had a 256 x 256 pixel image with 16-bit pixels, there are 2^(2^20) different images you could make, which is going to make any individual image's probability so tiny as to pretty much defy interpretation.

If we wanted to know $p(x|y)$, we'll use Bayes' Law to compute $p(x|y) = \frac{p(y|x)p(x)}{p(y)}$

On the other hand, $p(y|x)$ could be represented as a single-variable normal distribution representing our belief in $y$ given that you know $x$, which is the task that is usually more tractable, and thus we're usually more interested in solving.

  • $\begingroup$ So is the answer just because, we have the random variable $e^{(i)}$, that is a function of two other random variables $x^{(i)}$ and $y^{(i)}$. However, when we are trying to predict new labels, we are given $x^{(i)}$ (without $y^{(i)}$ of course, otherwise we would just return $y^{(i)}$ as the prediction). Since we are given $x^{(i)}$ then $p(e^{(i)})$ becomes $p(y^{(i)} | x^{(i)}$? Right? Its just because of what we are given when dealing with predictions, we are not usually given both x and y when trying to predict (we only have both when we train basically). $\endgroup$ Oct 26, 2014 at 18:02
  • $\begingroup$ Thnx that made sense to me! :D $\endgroup$ Oct 26, 2014 at 18:09
  • $\begingroup$ x|y will not in general be normally distributed. I.e. if y|x ~ Normal(x,1) and x ~ Exp(1) $\endgroup$
    – sega_sai
    Oct 24, 2017 at 13:22
  • $\begingroup$ what is exogenously fixed? $\endgroup$ Jul 1, 2019 at 23:42

The issue I was having is since $e^{(i)}$ is a r.v in terms of $x^{(i)}$ and $y^{(i)}$. i.e.

$$e^{(i)} = y^{(i)} - \theta^{T} x^{(i)}$$

Then if we have:

$$p_{e}(e^{(i)}) = \frac{1}{\sqrt{2\pi\sigma}}exp \left( \frac{-(y^{(i)} - \theta^Tx^{(i)})^2}{2\sigma^2} \right)$$

when should we favor $p(x^{(i)} \mid y^{(i)})$ vs $p(y^{(i)} \mid x^{(i)})$? (basically, it depends what we observe!)

Basically the answer ends up being simple. We are interested in modeling $p(y^{(i)} | x^{(i)})$ because we want to predict y given x. So mathematically, $p(x^{(i)} \mid y^{(i)})$ vs $p(x^{(i)} \mid y^{(i)})$ are extremely similar and related by $p_{e}(e^{(i)})$. However, they different in terms of what they have fixed fixed (i.e. what is observed or given). If x is given, then its fixed. So because we usually have x during our prediction phase, then we just use the form of the conditional distribution we need, i.e. we use:

$$p(y^{(i)} | x^{(i)})$$

because we are given $x^{(i)}$. We do know what $p(x^{(i)} \mid y^{(i)})$ distribution looks like but its not useful since we usually are not given the label y without knowing its corresponding x.

  • 2
    $\begingroup$ Isn't this exactly what @Glen_b was originally saying in the comments to your question? $\endgroup$
    – whuber
    Oct 29, 2014 at 15:11
  • $\begingroup$ When I read Glen's comment at the time,didn't make sense to me.For some reason the person that actually posted an answer to my question made me realize what I was confused about.However, though the answer didn't provide enough detail, because, if my former confused self would have read his answer, I am not sure if I could have figured it out. Therefore, I wrote an answer that was detailed enough that I think my former me (or any future confused person) could for sure understand if they had the same doubt. Basically, I wrote an answer that I thought could for sure clarify any confused person. $\endgroup$ Oct 30, 2014 at 2:46
  • $\begingroup$ Glen's comment could have been correct but didn't clarify my doubt. Now that I understand the issue, thanks to the answerer (the reason I provided him with the accepted and liked answer). My answer is just to make sure that if the first answer didn't make sense to someone, that my way of explaining it would. Glen's comment was/is too short and didn't emphasize enough (or emphasize in the correct way) what was confusing me. I believe my answer and the other answer do and therefore they exist. Sorry if I wasn't smart enough at the time $\endgroup$ Oct 30, 2014 at 2:46
  • $\begingroup$ Actually, I can point you to the exact sentence that made me clarify my doubt: "so we're rarely interested in guessing x directly based on what we know about y." From michael's first answerer was crucial for me to come to the realization I wrote above. The reason I accepted his answer and not mine. Does that make sense? I think if Glen would have provided an answer, I would have for sure gave him all the credit etc, but it was all thanks to michael! :D We can delete these comment if you desire after you read them since they are not related to the question really. $\endgroup$ Oct 30, 2014 at 2:51
  • $\begingroup$ Btw, thanks for making me read glens, mine and michael's answer! It was helpful to understand these concepts even more! :D (We can delete these comment if you desire after you read them since they are not related to the question really.) $\endgroup$ Oct 30, 2014 at 2:52

I think a very simple approach to understanding why $p(e^{(i)}) = p(y^{(i)}|x^{(i)})$ and not $p(x^{(i)}|y^{(i)})$, for the case you have shown, is that the units of error $e^{(i)}$ are same as the quantity $y^{(i)}$ and also the units of $\beta x^{(i)}$. Now, the normal equation is not really symmetric in $x^{(i)}, y^{(i)}$ because if $x^{(i)}$ is a vector of order $N$ and $y^{(i)}$ is a vector of order $M$ then in the equation presented above $e^{(i)}$ will be of order $M$ and not order $N$. For simple linear regression with variable $x^{(i)}$ of order $1$ and $y^{(i)}$ or order $1$, this is not directly evident. Furthermore, the $\sigma^2$ in the equation for a higher order case will be the covariance matrix $\Sigma$ which will of the order $M\times M$, so the symmetry will not be there anymore.

  • $\begingroup$ fantastic contribution! of course, the units of the error will match the units of the output because we usually assume the error is added to the output function $f(x)+\epsilon = y + \epsilon$. This definitively helps remember things quicker (even if its not a proof). (+1) $\endgroup$ Oct 24, 2017 at 15:00

I don't think this has anything to do with what you are trying to predict, rather I'd go with something along these lines: We have: $$p(e^{(i)}) = \frac{1}{\sqrt{2\pi\sigma^2}} exp(-\frac{(e^{(i)})^2}{2\sigma^2})$$ which can be re-written as: $$p(e^{(i)}) = \frac{1}{\sqrt{2\pi\sigma^2}} exp(-\frac{(y^{(i)}-\theta^Tx^{(i)})^2}{2\sigma^2})$$ Now the expression on the right hand side in the above equation can be interpreted as representing a probability distribution of some sort of relationship between x and y. Now if we say that $\theta^Tx^{(i)}$ represents the mean of this probability distribution, then because the mean is a constant value $x^{(i)}$ must be constant, i.e. in other words $x^{(i)}$ is fixed. Hence, this expression then equals $p(y^{(i)} \vert x^{(i)})$.

Note that $p(y^{(i)},x^{(i)})$ would imply that both $y^{(i)}$ and $x^{(i)}$ are variable and so neither $y^{(i)}$ nor $\theta^Tx^{(i)}$ in that expression would have a constant value. So none of them could be the mean.

Also note that we can write down $p(x^{(i)} \vert y^{(i)})$ but for that we'll need to first re-write the expression $y^{(i)} - \theta^Tx^{(i)}$ in the form $x^{(i)} - \alpha y^{(i)}$ where $\alpha$ is some constant dependent on $\theta$. In that case, $p(x^{(i)} \vert y^{(i)}) \sim \mathcal{N}(\alpha,\sigma^2)$. This is because the normal distribution is given by: $$p(z)_{z \in \mathcal{N}(\mu, \sigma^2)}= \frac{1}{\sqrt{2\pi\sigma^2}}exp(-\frac{(z-\mu)^2}{2\sigma^2})$$


I also find Andrew Ng's notes confusing because there is a subtle point that isn't explained. What they say is that the noise $\epsilon$ has Gaussian distribution. This ends up being essential. If you look at the equation of a Gaussian:

$$ Gau(x,y,\theta) = \frac{1}{\sqrt{2 \pi} \sigma} exp\left( \frac{(y - \theta^T x)^2}{2 \sigma^2}\right)$$

its just a function of 3 variables. When you have a value for all of the variables (i.e. its inputs are "fixed") it outputs some value. However, how you "fix" a value in probability matters a lot. If you "fix" a variable due to conditioning then it means that you do an integral and re-normalization, but simply fixing because you are querying what the probability of a certain value is observed does not change the form of the equation while condition does. In other words, they have (assume $\theta$ is not a r.v. nor bayesian for simplicity):

$$ p(\epsilon) = p(y - \theta^Tx) = p(x,y;\theta) = p(x \mid y; \theta) = p(y \mid x; \theta) = \frac{1}{\sqrt{2 \pi} \sigma} exp\left( \frac{(y - \theta^T x)^2}{2 \sigma^2}\right) $$

i.e. they are all literally (and in analytic form) the same equations and return the same numbers. However, this only happens because things are Gaussian. If the noise were something else this would not happen. i.e. what happens for them is:

$$ g(x,y;\theta) = p(x,y;\theta) = \frac{ p(x,y;\theta)}{ \int_y p(x,y; \theta) dy } = \frac{p(x,y;\theta)}{\int_x p(x,y ; \theta) dx} $$

which is not generally true (I only know it true for Gaussian distributions).

For example, notice that the distribution for the noise $\epsilon$ is just a function of three variables (2 r.v./random and 1 not r.v./random) $x,y$ and $\theta$. Depending on what type of condition and values $x,y$ have you will get different values. For example consider the following counter example I cooked up:

enter image description here

this example just shows that conditioning does indeed change things quite a bit. In fact, conditioning tells you which table to choose $X \mid Y$ or $Y \mid X$ or simply the joint $X,Y$. The value $\theta$ could hypothetically choose a different set of tables (obviously not shown). So the main points are:

  1. things work so nicely in Ng's example because things remain Gaussian because they started off Gaussian
  2. The way you fix things matters a lot. Just wondering what the probability is that you observe $Y=y$ is not the same as conditioning on the event $Y=y$ happening and then asking something else. Fixing due to conditioning changes the table, just fixing due to a query fixes a value but to does not change the which table your looking at.


Some other points I thought would be interesting is that in the context of MLE we are usually searching for some good value of $\theta$. Thus, we want to get the $\theta$ we care most according to our objective. In this case since in practice usually we observe $x$ and then want to predict $y$ it makes sense to take $\theta$ that optimizes such a value i.e:

$$ \theta \in arg \max_{ \theta } P( \cap^N_{n} Y = y_n \mid X=x_n ; \theta) $$

this matters because as I showed in the previous table, optimizing $p(x,y;\theta)$ or $p(x \mid y ; \theta)$ might be tractable (or not) but they are probably don't result in the same $\theta$'s, since they might be optimizing different equations. So at Glen_b said in the question its because:

Because you're trying to predict y from x ?

seems obvious but it seems they might even result in different estimators if your not careful.


Let me give my take on this issue.

In linear regression we assume the following model: $$ y= \theta_0x_0+\theta_1x_1+\theta_2x_2+...+\theta_nx_n + \epsilon \\ = \theta^Tx + \epsilon $$

where $y$ and $x$'s are assumed to be random variables (there is no assumption on their distribution) and $\epsilon$ is another random variable which is assumed to be normally distributed.

The aim is given some observations (or realizations) of the random variables ($y$ and $x$) to find the most likely values of the parameters ($\theta$s). Never forget this aim, to repeat, our aim is finding the most likely values of the parameters given some observations and the linear model assumption. (This method is known as maximum likelihood estimation (MLE) but I won't go into that, I think it is intuitively clear).

One complication here is that in MLE we normally write first the likelihood of the data. But here we can't since we do not know that distributions of the random variables $y$ and $x$. However, we can write the following conditional distribution:

$$ p(y | x; \theta) = \frac{1}{ \sqrt{2\pi\sigma^2 }}\exp\left(-\frac{(y-\theta^Tx)^2}{2\sigma ^2 }\right) $$

Coming to the OP's concern, can't we write it like the following:

$$ p(x | y; \theta) = \frac{1}{ \sqrt{2\pi\sigma^2 }}\exp\left(-\frac{(y-\theta^Tx)^2}{2\sigma ^2 }\right) $$

Yes this is also possible and the probability distribution will be the same. You can continue the derivation in the same way and estimate the most likely values of the parameters.

Whether we predict $y$ from $x$ or $x$ from $y$ is not important here. But as others have said since we usually use y to denote the target variable, we usually predict $y$ from $x$. To emphasize again, in this problem formulation our aim is not to predict $y$ from $x$, rather, our aim is to find the most likely values of the parameters given the data. Once you find the values you can of course predict $y$ given the $x$ values, but you can also do other things like predict one of the feature values ($x_i$) given $y$ and other feature values or try to interpret the strength of the relationship between x and y.

This is rather a late reply, but since this is a fundemantal method, I'm writing this reply in the hope that some people might find it useful.

  • $\begingroup$ curious, what do you think of my answer here stats.stackexchange.com/a/319674/28986 I think the fact both are the same is a mathematical coincidence in this specific case. $\endgroup$ Feb 17, 2023 at 19:00
  • $\begingroup$ Sorry, I could not follow your reasoning there, but my point is that we have to think about everything in the context of MLE, otherwise conditioning on x seems to be arbitrary and does not make sense. And I agree that there are logical gaps in Andrew Ng's notes. But in general this situation is everywhere in stats, authors take shortcuts which puzzles careful readers. $\endgroup$
    – Sanyo Mn
    Feb 19, 2023 at 10:56

the property of Normal distribution is If a random variable X has a normal distribution N(θ, σ2) a new random variable aX + b has a normal distribution N(aθ + b, a2σ2)

the response yi is a linear function of a set of features x:

y ( i ) = θ T x ( i ) + ϵ ( i )

Now, let us assume that all errors has a Normal Distribution N(0, σ2) Therefore, we can immediately know that yi will also follow a Normal Distribution N(θTx, σ2) by using the property we introduced at the beginning of the section:

enter image description here

  • $\begingroup$ but there is some conditioning going on that must keep it normal too, right? $\endgroup$ Sep 7, 2016 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.