Recursive proof of expectation of geometric distribution?If T ∼ G e o ( p ) ( 0 &lt; p &lt; 1) then E T = 1 / p is well known.Suppose that I toss a sequence of coins that come up heads with probability p. On average, how many coins do I have to toss before I see a heads?

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Recursive proof of expectation of geometric distribution?If   T  ∼      G    e    o    (  p  ) (  0  &amp;lt;  p  &amp;lt;  1) then       E    T  =  1      /    p is well known.Suppose that I toss a sequence of coins that come up heads with probability p. On average, how many coins do I have to toss before I see a heads?

Jovanni Salinas · Accepted Answer

Step 1We can condition on the first toss, let H denote the event that the first toss is a head.                    E        [        T        ]                            =        E        [        T                  |                H        ]        E        [        H        ]        +        E        [        T                  |                          H          c                ]        E        [                  H          c                ]                                          =        p        +        (        1        +        E        [        T        ]        )        (        1        −        p        )                                          =        1        +        (        1        −        p        )        E        [        T        ]            Step 2Solving for E[T] gives us       1    p  .Notice that   E  [  T      |        H    c    ]  =  1  +  E  [  T  ] as the first toss and the remaining tosses are independent.

Suppose that I toss a sequence of coins that come up heads with probability p. On average, how many coins do I have to toss before I see a heads?

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