Let $\left( X _ { k } \right) _ { k \in \mathbf{N} ^ { * } }$ be independent random variables with the same distribution given by: $$P \left( X _ { 1 } = - 1 \right) = P \left( X _ { 1 } = 1 \right) = \frac { 1 } { 2 }$$ For all $n \in \mathbf { N } ^ { * }$, we denote $S _ { n } = \sum _ { k = 1 } ^ { n } X _ { k }$. Conclude that: $$\forall n \in \mathbf { N } ^ { * } , \quad E \left( \left| S _ { 2 n } \right| \right) = E \left( \left| S _ { 2 n - 1 } \right| \right) = \frac { ( 2 n - 1 ) ! } { 2 ^ { 2 n - 2 } ( ( n - 1 ) ! ) ^ { 2 } }$$
Let $\left( X _ { k } \right) _ { k \in \mathbf{N} ^ { * } }$ be independent random variables with the same distribution given by:
$$P \left( X _ { 1 } = - 1 \right) = P \left( X _ { 1 } = 1 \right) = \frac { 1 } { 2 }$$
For all $n \in \mathbf { N } ^ { * }$, we denote $S _ { n } = \sum _ { k = 1 } ^ { n } X _ { k }$.
Conclude that:
$$\forall n \in \mathbf { N } ^ { * } , \quad E \left( \left| S _ { 2 n } \right| \right) = E \left( \left| S _ { 2 n - 1 } \right| \right) = \frac { ( 2 n - 1 ) ! } { 2 ^ { 2 n - 2 } ( ( n - 1 ) ! ) ^ { 2 } }$$