Tornados_MB.Rmd

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
### Marie Bonitz
### Übung 1: Tornados

## The 1,000-year tornado occurs at least onces in a thousand years with >60% probability.
**Verification of this statement:**

To calculate the probability of observing **at least one** event we have to determine the inverse probability of observing **none**.

```{r one occurance of 1000a event}
# set number of years
n <- 100
# set annual exceedence probability
p <- 1/n
# probability of no 1,000-year event in 1000 years
(1-p)^n
```
The result is the probability for no 1,000-year tornado to occur.

The probability for at least one tornado to occur is 63%. 
**Statement verified!**

## What can we say about the probabilities of observing multiple 1,000-year tornadoes in thousand years?

To find out the different probabilities for multiple events we have to simulate a time series.

```{r more than one occurance of 1000a event_1}
# generation of thousand random numbers
y <- runif(1000)
# counting all numbers >=0.999
(obs <- sum(y >= 0.999))
```
The result is the number of 1,000-year tornadoes that occured in this simulation. If we would generate new numbers and count them, another number of hurricanes could be our result.

Now we repeat this simulation 1000 times to calculate the precise probabilities.

```{r more than one occurance of 1000a event_2}
# generate 1000 time series and list it column-wise
y_long <- matrix(runif(1e6), ncol = 1000)
# compute the column sums
obs_long <- colSums(y_long >= 0.999)
```
The result ist a large matrix with 1 million elements.

For a better presentation of the distribution we plot the results.

```{r plot the result}
plot(table(obs_long), lwd = 10,
     main = "Simulated 1,000-year Tornadoes",
     col = "blue",
     xlab = "Tornado count",
     ylab = "Frequency",
     cex.lab = 1.5, cex.main = 1.5)
```
    
The probability to observe one 1,000-year tornado in 1000 years should usually be the highest. Though it is possible, that some distributions diverge from that.