0
$\begingroup$

I have one question regarding the estimation of an unknown variable.

Is there any way of estimating the value of a variable when you know its probability distribution? In this case I have a variable which is distributed on the interval (1, 2), with a 25% probability of 2, and uniformly distributed otherwise.

$\endgroup$
  • 1
    $\begingroup$ Cross-posted at statalist.org/forums/forum/general-stata-discussion/general/… Telling people about cross-posting is polite in any forum, and some have explicit policies about it. 1.625 is the mean of the distribution. There might be grounds for other prediction rules. but I can't see any here. $\endgroup$ – Nick Cox Nov 20 '16 at 11:37
  • $\begingroup$ Do you want to sample from that distribution? $\endgroup$ – mdewey Nov 20 '16 at 12:22
  • $\begingroup$ It's not really clear what you're asking. Can you explain more about what you're trying to achieve? $\endgroup$ – Glen_b Nov 21 '16 at 10:46
1
$\begingroup$

A random variable doesn't have a single unique value (unless it's degenerate, and your variable isn't). On the contrary, random variables exist to provide a mathematical formalism for something that might take on any of a variety of values.

Perhaps you mean to ask what the mean or median of this variable is. See the Wikipedia article "Central tendency".

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

I think that you should define a loss function and select an estimate that minimizes expected value of the loss for your distribution.

  • If you have quadratic loss $E (y - \hat{y})^2 \rightarrow \min$, then your $\hat{y}$ equals to the mean values of the distribution $E y$.
  • If you have $L1$ loss $E |y - \hat{y}| \rightarrow \min$, you shoud use median of the distribution as an estimate $\hat{y}$ that minimize the target loss
  • For loss function $E [y \ne \hat{y}]$ ($[x]$ is the indicator function that equals $1$ if $x = 1$ and $0$ otherwise) you should use the most probable values, $2$ in your case.
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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