# Tag Info

## Hot answers tagged distributions

6

The mean minimizing the root mean square error is often not the practical situation It is well known that the mean E(Y |X) minimizes the Root Mean Square Error (RMSE). You are right, the theoretical mean $E(Y |X)$ minimizes the root mean square error of a prediction (independent of the distribution). So if minimizing the mean square error of a prediction ...

5

This is laplace distribution, you can either use the CDF given in the wikipedia page or find a proof. If you want to integrate it yourself, the integral will be from $-\infty$ (not $0$) to $x$, and for the negative portion, you'll just substitute $|x|=-x$ and do the integration.

4

To understand their relation, you should go back to how $\sigma^2$ is defined. Recall that in the discrete case $$\sigma^2=Var(X)=E[(X-\mu)^2]$$ If you have have all observations in the population, you can calculate this expected value by the formula you first provided $$\sigma^2=\frac{1}{N}\sum^N_{i=1}(x_i-\mu)^2.$$ When $X$ instead is a random variable, ...

2

Partially answered in comments: Terms like "square-normal" usually emulate the model of the "log-normal" distribution: in the latter case, to say 𝑌 has a lognormal distribution mean the log of $𝑌$ is Normally distributed. Similarly, then, to say $𝑌$ has a "square-normal" distribution would ordinarily be understood as meaning ...

2

When the distribution (family) is known, and the sample size is large, it is generally difficult to find something better than maximum likelihood. Anyhow, to give more specific advice than that we would need to know the specifics. It is unclear why you think binning is more robust, generally it is just information loss. But you can still use maximum ...

2

Are your data "block maxima"? Such as, max annual values covering multiple years of time? Or something similar? If so, the the GEV is justified, but you might ask yourself if it is informative, since the GEV is so widely encompassing of samples from such a wide diversity of source processes. If you can narrow things down, that helps. So, if the ...

2

I'm pretty sure the expected answer to this problem involves an argument such as the one suggested by @whuber. However, because the values are specific, a grid search in R can be used to find an exact solution, as shown below. I assume $3$ is the gamma rate parameter, as in R. [There is no general agreement whether the second parameter of a gamma random ...

2

Let $Z$ be a random variable with distribution $P^Z$, meaning that for any measurable set $A$,$$\mathbb P(Z\in A)=P^Z(A)$$ Then, for any measurable transform $f$, $X=f(Z)$ is a random variable with distribution $P^X$ such that, for any measurable set $A$,$$P^X(A)=\mathbb P(X\in A)=\mathbb P(f(Z)\in A)=\mathbb P(Z\in f^{-1}(A))=P^Z(f^{-1}(A))$$ where f^{-1}(...

2

The answer is simpler than it seems. Though the sample mean in the simplest case, or least squares estimates in the multivariable predictor case provide unbiased estimates of the long-run mean, these estimates can be wrong or highly inefficient. In the case of a simple mean, i.e., when there are no predictors X, if the sample comes from a log-normal ...

1

You are doing hierarchical forecasting, specifically using the top-down method. That is, you first forecast the aggregate, then break it down. An alternative would be the bottom-up method, where we first forecast daily orders, then aggregate them up. In top-down forecasting, there are of course different ways of disaggregating. What you are doing is ...

1

First of all, it is a common misconception that a probability distribution only "exists" if it is of one of the forms found in the families of distributions that receive attention in academic literature and are therefore listed on resources like Wikipedia. Every discrete distribution obeying the rules of probability exists, regardless of whether ...

1

Here is one unnecessary way to approximate the answer to this question, using normal distributions. The expected number of cars $X$ is $E[X]=0.25 \cdot 0+0.5 \cdot 1+0.25 \cdot 2=1$. The standard deviation of $X$ is $SD[X]=\sqrt{(0-1)^2 \cdot 0.25+(1-1)^2 \cdot 0.5+(2-1)^2 \cdot 0.25}=\sqrt{0.5}=0.707$. Using convolutions we can sum 100 such normal ...

1

For actually generating $n$ such values, here are three approaches in R turning uniform pseudo-random variables into your distribution. The first is a simplified inversion of the CDF, putting its two parts together; the second essentially takes an exponential distribution and applies a random sign; and the third is essentially the difference between two ...

1

Rather than directly answer each point, I think it is better to explain how to think about these questions. I would disagree with even your answer to h). Just because the question states the number of pennies does not mean it has anything to do with flipping coins. For each scenario you should consider the distributions, and what they represent. For instance,...

1

If your data resembles as if sampled from a t-distribution rather than from a normal distribution, that means it has fatter tails, while it is still symmetric. You cannot say from this how the data was generated, there must be many possibilities. But it must be some process that for some reason tends to produce many atypical values outliers. One way to ...

1

I think I disagree with Ingolifs's answer. If you can justify, in this case on physical grounds, any one of the distributions, then that distribution is worthy of quite a bit of focus, including possibly for statistical hypothesis testing. I'm most familiar with the lognormal. If your physical system can result in data that can be envisioned as resulting ...

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