Statistics with R

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# Statistics with R
## Introduction to R for Actuarial Students

---

* Introduction to R for Actuarial Students

* CS1B Curriculum

* Introduction to R programming
* Fundamentals of Statistical Analysis
* Probability Distributions

* Question 2 - Random Number Generation
* Exam on basis of ***Base R***

---

### Reproducible Simulations

When you execute any R code in this question, make sure you run the entire R script 
including the line ***set.seed(1234)*** at the start.

---

Use the command <tt>set.seed(1234)</tt> to initialise the random number generator.

#### Exercises
1.  Generate a sample of size `$n = 180$` from an Exponential distribution 
with parameter `$\lambda  = 5$`.

2. Calculate the sample median of the sample in part 1.

3. Calculate the median of the Exponential distribution with `$\lambda  = 5$`.

4. Comment on your results in parts 2 and 3.

5.  The maximum likelihood estimator for `$\lambda$` is given by `$\lambda = 1/X$` where `$X$` is the sample 
mean. Estimate the parameter `$\lambda$`  using the maximum likelihood estimator and 
the sample in part 1

6. Generate another sample of size `$n = 180$` from an Exponential 
distribution with parameter `$\lambda  = 5$` and estimate the parameter `$\lambda$`  using 
the maximum likelihood estimator and the new sample.

7. Comment on the estimated values in parts 5 and 6

---

### Set Up

```r
set.seed(1234)

lambda <- 5

samplesize <- 180
```

---

#### Exercise 1

Generate a sample of size `$n = 180$` from an Exponential distribution 
with parameter `$\lambda  = 5$`.

```r
x = rexp(samplesize,lambda) 
```

```r
head(x)
```

```
## [1] 0.500351721 0.049351777 0.001316391 0.348549218 0.077436517 0.017989934
```

#### Alternative Approach

* Using Inverse Transform Sampling

```r
x=-(1/5)*log(runif(samplesize))
```

---

#### Exercise 2

Calculate the sample median of the sample in part 1.

```r
m = median(x) 
m 
```

```
## [1] 0.1306305
```

---

#### Exercise 3

Calculate the median of the Exponential distribution with `$\lambda  = 5$`.

```r
mq = qexp(0.5,lambda)
mq 
```

```
## [1] 0.1386294
```

Remark: This is the analytical result, i.e. the population median

---

#### Exercise 4

Comment on your results in parts 2 and 3.

* The sample median in part 2 is an estimator for the true median in part 3 based
on the sample in part 1

* The results in parts 2 and 3 are not equal but similar

---

#### Exercise 5
The maximum likelihood estimator for `$\lambda$` is given by `$\lambda = 1/X$` where `$X$` is the sample 
mean. Estimate the parameter `$\lambda$`  using the maximum likelihood estimator and 
the sample in part 1

* For the parameter of an exponential distribution the MLE is the inverse of the
sample mean

```r
l = 1/mean(x) 
l 
```

```
## [1] 5.259869
```

Compare to `$\lambda$`

```r
lambda
```

```
## [1] 5
```
---

#### Exercise 6

Generate another sample of size n = 180 from an Exponential distribution with parameter `$\lambda  = 5$` and estimate the parameter `$\lambda$` using the maximum likelihood estimator and the new sample.

```r
x2 <- rexp(samplesize,lambda)

1/mean(x2) 
```

```
## [1] 4.894128
```

---

#### Exercise 7

Comment on the estimated values in parts 5 and 6

* The estimated values are both close to the true value ($\lambda = 5$) but not identical. 
* The actual value of the estimator is a random variable and the estimated value depends on the actual sample

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