Question

Prove Proposition$4.4$by mathematical induction.

Short answer

To prove the statement for$k+1,$considering first $k$events as one, use Proposition$4.3.$, and then the presumption for$k$twice.

Step-by-step solution

Given Information.

Given that Proposition$4.4$.

explanation.

We will apply mathematical induction to the number of sets. For$n=2$, see Proposition $4.3.$Now assume for $n=k$we have. $P\left(E_1\cup E_2\cup \dots \cup E_k\right)=\sum _{i=1}^{k}P\left(E_i\right)-\sum _{i_1<i_2}P\left(E_{i_1}{E}_{i_2}\right)+\dots +(-1{)}^{r+1}\sum _{i_1<\dots <i_r}P\left(E_{i_1}\dots E_{i_r}\right)+\dots +(-1{)}^{k+1}P\left(E_1\dots E_k\right)$

For $n=k+1.$Set $E=E_1\cup \dots \cup E_k$and apply Proposition $4.3.$Then we get

$P\left(E_1\cup E_2\cup \dots \cup E_k\cup E_{k+1}\right)=P\left(E\right)+P\left(E_{k+1}\right)-P\left(E{E}_{k+1}\right)$

Using the induction hypothesis we get

$P\left(E_1\cup E_2\cup \dots \cup E_{k+1}\right)=P\left(E_{k+1}\right)+\sum _{i=1}^{k}P\left(E_i\right)-\sum _{i_1<i_2}P\left(E_{i_1}{E}_{i_2}\right)+\dots +(-1{)}^{r+1}\sum _{i_1<\dots <i_r}P\left(E_{i_1}\dots E_{i_r}\right)+\dots +(-1{)}^{k+1}P\left(E_1\dots E_k\right)-P\left(E{E}_{k+1}\right)$

Now observe that

$$\begin{gathered} P\left(E{E}_{k+1}\right)=P\left(\left(E_1\cup \dots \cup E_k\right)E_{k+1}\right)=P\left(E_1{E}_{k+1}\cup \dots \cup E_k{E}_{k+1}\right) \\ =\sum _{i=1}^{k}P\left(E_i{E}_{k+1}\right)-\sum _{i_1<i_2}P\left(E_{i_1}{E}_{i_2}{E}_{k+1}\right)+\dots +(-1{)}^{r+1}\sum _{i_1<\dots <i_r}P\left(E_{i_1}\dots E_{i_r}{E}_{k+1}\right)+\dots +(-1{)}^{k+1}P\left(E_1\dots E_k{E}_{k+1}\right) \end{gathered}$$

Combining this with the identity above we get

$P\left(E_1\cup E_2\cup \dots \cup E_{k+1}\right)=\sum _{i=1}^{k+1}P\left(E_i\right)-\sum _{i_1<i_2}P\left(E_{i_1}{E}_{i_2}\right)+\dots +(-1{)}^{r+1}\sum _{i_1<\dots <i_r}P\left(E_{i_1}\dots E_{i_r}\right)+\dots +(-1{)}^{k+2}P\left(E_1\dots E_{k+1}\right)$

Hence we get the desired identity by mathematical induction.