Lab 4: Reinsurance. Data frames

1 Reinsurance

For the next task, you will need to use function pmax. If a is a vector \((a_1,\ldots,a_m)\) and x is a number, then usual max(a,x) returns \[ \max\{a_1,\ldots,a_m,x\} \] that is a number, whereas pmax(a,x) returns \[ \bigl(\max\{a_1,x\},\ldots,\max\{a_m,x\}) \] that is a vector.

1.1

Set seed \(234\). Simulate \(15000\) values for a reinsurer where (aggregated) claims have a compound Poisson distribution with parameter \(900\) and the individual claims have log-normal distribution \(logN(7,0.3^2)\) under individual excess of loss with retention \(1700\). Find the mean and the standard deviation of the obtained data, i.e. using the notations from the lecture notes, estimate \(\E(S_R)\) for

\[ S_R=Z_1+\ldots+Z_N \]

from the sample, and also estimate \(\sigma(S_R)\) from the sample.

Check the answer.

[1] 16640.800  2872.169

1.2

We can see, using formulas from the lecture notes, that in the assumption of the previous task \[ \E(S) = \E(N)\E(X)=900\cdot e^{7+0.5\cdot 0.3^2}=1032398 \] (you can also check that the mean of the sample: for \(S\) (not \(S_R\)), with the seed \(234\) it would be \(1032358\), pretty close). Suppose now that the reinsurer applies the aggregated excess \(M\), i.e. the insurer pays \[ S_I=\begin{cases} S, & \text{if } S \leq M,\\ M, & \text{if } S > M \end{cases} \] and the reinsurer pays \[ S_R=S-S_I=\begin{cases} 0, & \text{if } S \leq M,\\ S-M, & \text{if } S > M. \end{cases} \]

Let \(M=\E(S)=1032398\). Modify the previous code and find \(\E(S_R)\), the average paid by the reinsurer under this agreement.

Check the answer.

[1] 14297.37

2 How to create data frames

  • Data frame is a table with data.

  • Each column of a data frame should contain the data of the same type (e.g. only numeric or only characters), however, different columns may have different types:

df <- data.frame(x = 101:103, y = c('a', 'b', 'c'))
df
    x y
1 101 a
2 102 b
3 103 c

  • Here we create a data frame df with two columns: column x contains \(3\) numbers: \(101\), \(102\), \(103\) (see topic Patterns of Lab 1) and column y contains three strings “a”, “b”, and “c”.

  • On the left, you can also see numbers \(1,2,3\), these are the indexes which number the rows of the frame.

  • We will mainly need access to the columns of a data frame. The command has the following simple structure DataFrameName$ColumnName, e.g.

df$x
[1] 101 102 103

Stress that the output is a vector. Similarly,

df$y
[1] "a" "b" "c"
  • The data frame may be very large, and we may need to see its structure without showing the whole data frame. For this we can use function head(...):
x <- 101:10**3
y <- 2*x + 7
df <- data.frame(x, y)
head(df)
    x   y
1 101 209
2 102 211
3 103 213
4 104 215
5 105 217
6 106 219
  • By default, head shows \(6\) rows. One can request more:
head(df,10)
     x   y
1  101 209
2  102 211
3  103 213
4  104 215
5  105 217
6  106 219
7  107 221
8  108 223
9  109 225
10 110 227
  • Try commands ncol(df), nrow(df), dim(df) to see their outputs.

2.1

Find the sum of all values in the column y of this data frame df.

Check the answer:

[1] 997200

3 How to import data frames

  • Often we need to work with external data stored in some files. A popular format for storing text data (that includes numbers) is CSV-format (stands for “fomma separated values”).

3.1

Do the following steps:

  1. Download file numbers.csv and save it in a directory (folder) on your computer (usually it’s “Documents” or “Downloads”).

  2. Save your R-file your are working with in R-Studio to the same directory (folder).

  3. Click with right mouse button on the file name at the top of R-Studio window and choose “Set working directory”.

OR (instead of 3)

  • Go to menu “Session”, choose “Set working directory”, and then choose “To source file location”

If you did 1-3 correctly, in “Files” tab on the (down) right part of R-Studio you will see your R-file and the downloaded CSV-file in the same directory.

  1. Load the saved file by the command
df <- read.csv("numbers.csv")

Check yourself:

head(df)
           x        y
1 1.98193075 8.554741
2 0.08423538 4.008477
3 0.33576991 4.920004
4 0.44245007 4.861317
5 0.09377962 4.023355
6 0.42974243 5.109215

3.2

It is given that the data in column x of “numbers.csv” file are independent values of a random variable \(X\sim Exp(\lambda)\). Find the estimate \(\hat{\lambda}\) for the parameter \(\lambda\) using the method of moments (see Lecture Notes).

Check your answer:

[1] 1.152757

4 How to filter data frames

  • Suppose we want to select only those rows of the last data frame where the value of y is bigger than 5. This condition then has the form: df$y > 5. Since df$y is a vector, the result of df$y > 5 will be also a vector of TRUE and FALSE:
my_condition <- df$y > 5
my_condition[1:6]
[1]  TRUE FALSE FALSE FALSE FALSE  TRUE

i.e. the first, second, forth, and sixth values of y are bigger than \(5\) (and of course many others, we are just looking here for the first \(6\)). The length of my_condition is the same as the number of rows in df:

length(my_condition) == nrow(df)
[1] TRUE
  • The command df[my_condition,] (notice the comma!!!) will return the data frame which contain only those rows from df where \(y>5\):
df_cond <- df[my_condition,]
head(df_cond)
           x        y
1  1.9819307 8.554741
6  0.4297424 5.109215
7  1.1831351 6.374403
8  0.2664600 5.516452
11 0.3309978 5.333065
13 0.7291787 5.758398
  • Since df_cond is a data frame (and the same column names are inheritted from df), we can access its columns by df_cond$x and df_cond$y.

4.1

Find the average of all values of y from df such that the corresponding values of x are larger than \(1\).

Check your answer:

[1] 7.687361

5 Linear models

  • Sometimes we may believe that two sets of data are related to each other. The simplest relation is linear, like \(y=kx+b\).

  • If one has sets of data \(x=(x_1,\ldots,x_n)\) and \(y=(y_1,\ldots,y_n)\), one can find \(k\) and \(b\) which would be the best possible to make $y_ik x_i +b $ for all \(i\).

  • Usually, it is done using the linear regression method which chooses \(k\) and \(b\) so that

\[ \sum_{i=1}^n(y_i - (kx_i+b))^2 \]

would be smallest possible (for the given \(x\) and \(y\)).

  • In the data frame above, we may suspect that \(y\) and \(x\) are linearly related:

  • The corresponding R-code to find optimal \(k\) and \(b\) is the following:
lm(y ~ x, data = df)

Call:
lm(formula = y ~ x, data = df)

Coefficients:
(Intercept)            x  
      3.933        2.071

  • Here data is the name of the parameter of function lm (stands for “linear model”), and df is the data frame which contains columns x and y. In other words, if columns of df would had names colX and colY, we would write lm(colY ~ colX, data = df).

  • The (Intercept) corresponds to \(b\), and value of x here is \(k\). One can access the obtained values:

ans <- coef(lm(y ~ x, data = df))
ans
(Intercept)           x 
   3.932877    2.070541

and then ans[1] is \(b\) and ans[2] is \(k\).

5.1

Plot on the same diagram the scatter plot \(y\) against \(x\) as it is shown above, but in blue color, and plot the line \(y=kx+b\) with the found values of \(k\) and \(b\) (do not copy-paste, use coef function) in red colour. (See Lab 3 for plotting two graphs on one diagram).

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