Show that the cdf for a geometric random variable is given by FX(t) = P(X ≤ t) = 1 − (1 − p)^[t], where [t] denotes the greatest integer less than or equal to t, t ≥ 0.
Added by Manuel S.
Step 1
The probability mass function (pmf) of a geometric random variable is given by: P(X = k) = (1 - p)^(k-1) * p, for k = 1, 2, 3, ... Now, we want to find the cumulative distribution function (cdf) F_X(t), which is the probability that X is less than or equal to Show more…
Show all steps
Close
Your feedback will help us improve your experience
Jacob Fry and 68 other Intro Stats / AP Statistics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Show that if the random variable $X$ has the geometric distribution with parameter $p,$ and $j$ is a positive integer, then $p(X \geq j)=(1-p)^{j-1} .$
Discrete Probability
Expected Value and Variance
Let $X$ be a random variable of the discrete type with pmf $p(x)$ that is positive on the nonnegative integers and is equal to zero elsewhere. Show that $$ E(X)=\sum_{x=0}^{\infty}[1-F(x)] $$ where $F(x)$ is the cdf of $X$.
Probability and Distributions
Some Special Expectations
Let $X$ be a random variable with mgf $M(t),-h<t<h$. Prove that $$ P(X \geq a) \leq e^{-a t} M(t), \quad 0<t<h $$ and that $$ P(X \leq a) \leq e^{-a t} M(t), \quad-h<t<0 $$ Hint: Let $u(x)=e^{t x}$ and $c=e^{t a}$ in Theorem 1.10.2. Note: These results imply that $P(X \geq a)$ and $P(X \leq a)$ are less than or equal to their respective least upper bounds for $e^{-a t} M(t)$ when $0<t<h$ and when $-\bar{h}<t<0$.
Important Inequalities
Recommended Textbooks
Elementary Statistics a Step by Step Approach
The Practice of Statistics for AP
Introductory Statistics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD