Statistical Analysis Stack Exchange
Statistical Analysis Stack Exchange

Questions tagged [geometric-distribution]

The geometric distribution is a discrete (count) distribution, where the probability of each count is a constant proportion of the next lower count. An example is 'the number of coin tosses until the first head'.

The geometric distribution is a discrete distribution. The probability of each value after the first is a constant proportion of the next lower value.

The geometric distribution is a special case of the negative binomial. It is the probability of a particular kind of count in a set of Bernoulli trials - and may be defined either as the number of trials or the number of failures to the first success (the count thereby being at least 1 or at least 0 respectively). An example is 'the number of coin tosses until the first head'.

See also: https://en.wikipedia.org/wiki/Geometric_distribution

References:
Wikipedia: Geometric distribution
Mathworld: Geometric Distribution

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Expected number of tosses till first head comes up

Suppose that a fair coin is tossed repeatedly until a head is obtained for the first time. What is the expected number of tosses that will be required? What is the expected number of tails that will be obtained before the first head is obtained?
nicole900
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Probability of $X_1 \geq X_2$

Suppose $X_1$ and $X_2$ are independent geometric random variables with parameter $p$. What is the probability that $X_1 \geq X_2$? I am confused about this question because we aren't told anything about $X_1$ and $X_2$ other than they are…
IrCa
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Fisher's score function has mean zero - what does that even mean?

I'm trying to follow the princeton review of likelihood theory. They define Fisher’s score function as The first derivative of the log-likelihood function, and they say that the score is a random vector. E.g for the Geometric distribution: $$ u(\pi)…
ihadanny
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Ms. A selects a number $X$ randomly from the uniform distribution on $[0, 1]$. Then Mr. B repeatedly, and independently, draws numbers

Ms. A selects a number $X$ randomly from the uniform distribution on $[0, 1]$. Then Mr. B repeatedly, and independently, draws numbers $Y_1, Y_2, ...$ from the uniform distribution on $[0, 1]$, until he gets a number larger than $\frac{X}{2}$, then…
Shreya Bhandari
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Estimating the parameter of a geometric distribution from a single sample

I was surprised not to find anything about this with Google. Consider a geometric distribution with $\text{Pr}[X=k]=(1-p)^{k-1}p$, so the mean is $\sum_{k=1}^\infty k\,\text{Pr}[X=k]=\frac{1}{p}$. Now assume we observe a single outcome (number of…
Tarvoc
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Strange connection between Bernouilli, Uniform and Geometric distributions

Final update on 11/29/2019: I have worked on this a bit more, and wrote an article summarizing all the main findings. You can read it here. Let us consider $Z = X_1 + X_1 X_2 + X_1 X_2 X_3 +\cdots$ where the $X_i$'s are i.i.d. I proved in my…
Vincent Granville
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Confidence Interval on the Geometric Distribution Expected Value?

If we are told that some random variable $X$ follows a Geometric distribution, with $Pr(X =1) = p$. The sample has observed values between $1$ and $N$. We know that $E(X) = 1/p$ My question is: Can we construct a confidence interval for the mean?
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How to bound a probability with Chernoff's inequality?

In my class, we were given Chernoff's inequality as $$P(X\le -t) \le e^{(-(\lambda t - \log( E(e^{-\lambda x}))))}$$ $$P(X\ge -t) \le e^{(-(\lambda t - \log( E(e^{\lambda x}))))}$$ It says that to find the best upper bound, we must find the best…
nomo
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Show that for a Geometric distribution, the probability generating function is given by $\frac{ps}{1-qs}$, $q=1-p$

Suppose that $X$ has a geometric distribution with probability mass function $P(X=x) = q^{i-1}p$, $i=1,2,...$ and $q=1-p$ Show that its probability generating function is given by $ \pi(s)=\frac{ps}{1-qs}$. Hence show that $E(x)=\frac{1}{p}$ and…
JackReacher
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A description of the mean of the Geometric Distribution - is it unorthodox or just incorrect?

I have a homework assignment where I'm asked to propose an estimator for the mean of a geometric random variable. This seemed simple enough, given that I've always understood the mean of the geometric distribution to be $1\over{p}$. The directions,…
jmars
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Geometric distribution with random, varying success probability

I'm really sorry if this question is too basic, but I've been looking for a while and haven't been able to find a convincing response. My statistics background is rather poor. Geometric distribution is defined as the probability distribution of the…
Rafael
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Convolution of random variables: unimodality of the likelihood function

Let $X_1, X_2,...X_k$ be random independent variables, each $X_i$ drawn from a Geometric distribution $\mathcal{G}(p_i)$, and let its convolution, or sum, be $Y = \sum_{i=1}^k X_i$. The likelihood function of each $X_i$ with respect to $p_i$ is…
alberto
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Geometric distribution without replacement

On an attempt to solve this problem I've managed to reduce it to finding the expected number of white balls picked until one black ball is observed (let's call that value $v$). Except that, unlike the geometric distribution, this needs to be done…
Mohammad Alaggan
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How to sample value from Symmetric Geometric Distribution

How can I sample a value from a symmetric geometric distribution, as defined in this link. There, the density of the symmetric geometric proposal distribution is given by \begin{equation*} f(\theta;p_g)\propto \frac{p_g (1 - p_g)^{|\theta|}}{2(1-…
Cesar
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If $Y \sim geometric(P)$ and $P \sim \mathcal B(2, 1)$ how to compute $E(Y)$ and marginal pmf of $Y$?

$$Y \sim Geometric(P)\\ P \sim \mathcal B(2, 1)$$ I'm trying to compute $E[Y]$ without finding marginal distribution of $Y$. I need some hints here. I also need to find the pmf of $Y$. My approach is as follows: Integrate from 0, 1 over $p:…
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