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A clever solution to find the expected value of a geometric r.v. is those employed in this video lecture of the MITx course "Introduction to Probability: Part 1 - The Fundamentals" (by the way, an extremely enjoyable course) and based on (a) the memoryless property of the geometric r.v. and (b) the total expectation theorem.
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Geometric Distribution - Memoryless property - geometric series. Related. 3. prove that any positive ...
To calculate this, you just sum the geometric series with first term $(1-p)^x p$ and ratio $1-p$, so we have $$ P(X>x) = \frac{(1-p)^x p}{1-(1-p)} = (1-p)^x, $$ as before. Share Cite
Regrettably, there are two distributions that are called geometric [1], the classical one, taking values in $1,2,\ldots$ and the shifted variant that takes values in $0,1,2,\ldots$.
Formula: Let $|q|<1$ then we have $$(\star) \ \ \sum_{k=1}^{\infty} q^k = \frac{q}{1-q}.$$ We use this fact for the calculations of MGF $$\mathbb{E}[e^{tX}] = \frac{p ...
A geometric random variable X has the memoryless property if for all nonnegative integers s and t , the following relation holds .
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This is for a geometric($\theta$) distribution. I am stuck on calculating the Fisher Information, which is ...