expected value of sample from a skewed distribution Let us say a market's prices are distributed with a skew.
Given a sample of 10000 products, I can estimate the median, arithmetic mean and weighted average or any other statistic.
Let us say I draw 100 products from this distribution, what would be their total expected/average value? I cannot use arithmetic mean * 100 or can I? Should I use median * 100? Or should one use 100 * weighted average?
I am sorry I am still confused about this, despite reading lots of related posts here and elsewhere. Any clarification would be very much appreciated please. Thanks!
PS:
Given Dave's answer and code below, I tried to simulate my situation. Basically I assumed a lognormal distribution to get a skew. I obtain some averages from a larger sample. I want to see which average to use to best estimate the total of N randomly sampled items. It turn out that, according to this (correct?) simulation the arithmetic mean is the better average in my situation. I guess this is some empirical evidence against my gut feeling that the median is the better/more representative average in my situation ...
p1 <- log(10)
p2 <- log(2)

# obtain estimates/averages from large sample from "the truth"
x <- rlnorm(1000000, p1, p2)
plot(density(x))
median <- median(x)
mean <- mean(x)

set.seed(2021)
N <- 100 # sample size
R <- 10000 # how many times to draw samples of size N
mean_estimate_se <- median_estimate_se <- rep(NA, R)
for (i in 1:R){
    
    x <- rlnorm(N, p1, p2) 
    
    sum <- sum(x)
    
    mean_estimate_se[i] <- ((mean * N) - sum)^2
    median_estimate_se[i] <- ((median * N) - sum)^2
    
}

paste0("Error using arithmetic mean: ", round(mean(mean_estimate_se), 2))
paste0("Error using median: ", round(mean(median_estimate_se), 2))

round(mean(median_estimate_se) / mean(mean_estimate_se), 2)


 A: Whether the distribution is skewed or not, the usual sample mean is always an unbiased estimator for the population mean.$^{\dagger}$
$$
\mathbb E\bigg[\bar X\bigg] = \mathbb E\Bigg[\frac{1}{n}\sum_{i = 1}^nX_i\Bigg]\\
= \frac{1}{n}\sum_{i = 1}^n \bigg( \mathbb E[X_i]\bigg) 
=
\frac{1}{n} \sum_{i = 1}^n \mu = \frac{1}{n}n\mu = \mu
$$
The usual sample mean is a perfectly defendable way to estimate the population mean.
Should gets into lots of fascinating ideas related to constructing estimators with desirable properties but might be a bit beyond where you want to operate. Depending on the particulars of the distribution, you might find the median to be a better estimator of the mean, perhaps based on the mean squared errors of those estimators.
$^{\dagger}$There's this annoying issues where the population mean might not exist. In that case, this calculation does not make sense, but you are trying to estimate a value that does not exist, which also does not make sense.
EDIT
In your comment, you mention that median might be a preferable estimator for a skewed distribution. Let's put that to the test in a simulation study.
set.seed(2021)
N <- 100 # sample size
R <- 1000 # how many times to draw samples of size N
means <- medians <- rep(NA, R)
for (i in 1:R){
    parameter <- 13 # the parameter of the distribution
    x <- rchisq(N, parameter) # this has a mean of parameter
    
    means[i] <- mean(x)
    medians[i] <- median(x)
    
    }

mean((means - parameter)^2)
mean((medians - parameter)^2)

I get an MSE of the sample mean of $0.25775941187491$ and an MSE of the sample median of $0.845599781948668$, more than three times as big. For this $\chi^2_3$ distribution, at least in terms of the common MSE metric, the mean is a vastly superior estimator, despite the skewness. As noted in the comments, other distributions, such as $t_3$, will give the opposite conclusion (despite the symmetry of $t_3$).
(I had trouble coming up with a skewed distribution that gave superior performance of the empirical median, and I welcome suggestions in the comments. For superiority of the empirical mean for a distribution that lacks skewness, look no further than the normal distribution. (Noted in the comments, absolute value of a low-degree-of-freedom t-distribution should do the trick.))
