Is the Jackknife estimation better than Maximum Likelihood Estimator? I'm trying to estimate distribution parameters with Maximum Likelihood Estimator (MLE) and Jackknife estimator based on it. The estimation statistic is mean. 
Jackknife estimator is considered to be better than MLE, but somehow I get the same numbers. 
What am I doing wrong?
I'm using the following R code:
# Function to return sample mean as estimator
MME_estimate_lambda <- function(vec){
    return(mean(vec))
}

# Function to perform Jackknife estimation
Jacknife_estimate_lambda <- function(vec){
    n <- length(vec)
    estimators <- sapply(seq(n),function(i) MME_estimate_lambda(vec[-i]))
    return(n*MME_estimate_lambda(vec) - (n-1)*sum(estimators)/n)
}

set.seed(123)
mysample <- rexp(50,1) # sample from Exponential distribution
MME_estimate_lambda(mysample)
Jacknife_estimate_lambda(mysample) # Both estimators are the same

 A: You are not doing anything wrong, the are suppossed to give the same number (in this simple setting):
# the true mean of y is 0
y <- rnorm(1000)

# The MLE estimator is simply the sample mean:
MLE <- mean(y)

# The Jackknife estimator is:  
require(bootstrap)
JKE <- mean(jackknife(y, mean)$jack.values)

# Both estimator's results in:
# 0.009662955
# Exatcly the same number.

# However the estimated standard error is different:
JKE_se <- jackknife(y, mean)$jack.se
# 0.0307548
MLE_se <- sqrt(var(y) * (length(y)-1) * 1/length(y))
# 0.9725519

Oops! The estimated sd for the jackknife is way off. There is general lesson here, provided you are willing to asumme the distributional assumptions imposed by MLE you cannot do any better. The nice thing about the jackknife, and orther resample methods, is that they do not impose such rigorous assumptions.
Of course today, with modern computers, there isnt really a reason to rely on the jackknife, instead you would almost always be better of using a simple bootstrap - because you learn something about the sampling distribution. 
A: Comparing maximum likelihood estimation (MLE) and the jackknife is like comparing apples and oranges. These techniques accomplish different things.

*

*Maximum likelihood estimation is a method for estimating parameters and fitting models. Since it's based on maximizing the likelihood function, it's a model-based algorithm.

*The jackknife is a method for estimating the bias and variance of estimators. It doesn't make distributional assumptions about the data generating process, so it's nonparametric.

Let's highlight the difference between estimating a parameter and estimating the variance of that estimator. I'll use @Repmat's example.
Let $x_1,\ldots,x_n$ be independent draws from a population with mean $\mu$. The sample average $\bar{x} = \sum_{i=1}^nx_i/n$ is an estimator of the mean $\mu$.
How accurately does $\bar{x}$ estimate $\mu$? In other words, what is the variance of $\bar{x}$?
The jackknife's answer is:
$$
\begin{aligned}
\widehat{\operatorname{var}}_{\text{jack}} &=
\frac{n-1}{n}\sum_{i=1}^n\left(\bar{x}_{(i)} - \bar{x}_{(.)}\right)^2
\end{aligned}
$$
where the $\bar{x}_{(i)}$ are the jackknife values:
$$
\begin{aligned}
\bar{x}_{(i)} = \sum_{j\neq i}\frac{x_j}{n-1}
\end{aligned}
$$
and $\bar{x}_{(.)}$ is the average of the $\bar{x}_{(i)}$s.
Now let's assume further that the $x_i$s are independent draws from a Normal distribution with mean $\mu$ and variance $\sigma^2$. Since we've made a distributional assumption, we can use maximum likelihood to estimate $\mu$ and $\sigma^2$.
$$
\begin{aligned}
\hat{\mu}_{\text{MLE}} = \bar{x}, \quad \hat{\sigma}^2_{\text{MLE}} = \sum_{i=1}^n\frac{(x_i - \bar{x})^2}{n}
\end{aligned}
$$
In the Normal model, how accurate is $\bar{x}$ as an estimate of $\mu$? We start by computing the variance of $\bar{x}$ in terms of $\sigma^2$.
\begin{align}
\operatorname{var}(\bar{x})
= \sum_{i=1}^n\frac{\operatorname{var}(x_i)}{n^2}
= \frac{\sigma^2}{n}
\end{align}
And since we don't actually know $\sigma^2$, we use the plug-in principle and substitute $\hat{\sigma}^2$ for $\sigma^2$.
To sum up, we have an estimator $\bar{x}$ of the mean $\mu$ and two estimates of the variance of that estimator, one derived with the jackknife and the other with maximum likelihood. Let's calculate them on some data.
set.seed(1234)

# We simulate some normal data but the jackknife makes no distributional assumptions.
n <- 1000
x <- rnorm(n)

# The MLE estimate of the population mean is the sample mean:
xbar <- mean(x)

# The MLE estimate of the population variance:
s2 <- sum((x - xbar)^2) / n

# The plug-in estimate of the variance of the sample mean:
s2 / n
#> [1] 0.0009936879

Now use the jackknife implementation in the bootstrap package.
library("bootstrap")

# The jackknife estimate of the variance of the sample mean:
(jackknife(x, mean)$jack.se)^2
#> [1] 0.0009946825

Okay. So the jackknife has a slightly bigger estimate for the variance of $\bar{x}$. Does it mean that it's better? The MLE estimate of $\sigma^2$ is biased by a factor of $n/(n-1)$: $\widehat{\sigma}^2_{\text{MLE}}$ is a little too small on average, so when we plug it in for $\sigma^2$ we underestimate the variance of $\bar{x}$. The jackknife estimate is corrected for that bias, hence it's better in a certain sense. For large samples the difference is negligible.
PS: With some algebra you can show that the difference between the MLE and the jackknife estimates of the variance of $\bar{x}$ is a factor of $n/(n-1)$.
PPS: What @Repmap calls MLE_se is the MLE estimate $\hat{\sigma}$ of the population standard deviation $\sigma$. Not an estimate of the standard error $\widehat{\operatorname{se}}$ of the sample mean $\bar{x}$.
