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I have a discrete random variable $X$ with probability function $$p_i=P(X=x_i), i=1,2,\ldots$$ such that $\sum_{i=1}^{\infty}p_i=1$. I need to generate $X^*$ such that $X^*$ has the same distribution as $X$. We generated $X^*$ in the following way: $$\begin{align} b_0 &= 0 \\ b_1 &= p_1 \\ b_2 &= p_1 + p_2 \\ b_j &= p_1 + p_2 + \ldots + p_j \\ b_{\infty} &= \sum_{i}p_i = 1 \end{align}$$ and generating $R\sim U[0,1]$, the discrete uniform distribution over $U[0,1]$, if $b_{j-1} < R \leq b_j$. Then we let $X^* = x_j$ and we have to show that $X^*$ has the same distribution as $X$. So to show this, I have been trying to show that $P(X^*=x_j)=P(X=x_j)=p_j$. I have that $X^*=x_j$ if $b_{j-1} < R \leq b_j$, so I rewrite $P(X^*=x_j)$ as $$ \begin{align} P(b_{j-1} < R \leq b_j) &= P(R \leq b_j) - P(R<b_{j-1}) \\ &=F_R(b_j) - (F_R(b_{j-1}) - f_R(b_{j-1}))\\ &=\frac{\lfloor{b_j}\rfloor + 1}{2} - (\frac{\lfloor{b_{j-1}}\rfloor + 1}{2} - \frac{1}{2}) \\ &=\frac{\lfloor{b_j}\rfloor - \lfloor{b_{j-1}}\rfloor + 1}{2} \end{align}$$ I'm not sure where to go from here, or if I'm even going in the right direction.
Probability Distributions > Uniform Distribution Contents:
1. What is a Uniform Distribution?A uniform distribution, also called a rectangular distribution, is a probability distribution that has constant probability. This distribution is defined by two parameters, a and b:
The distribution is written as U(a, b). The following graph shows the distribution with a = 1 and b = 3: Like all probability distributions for continuous random variables, the area under the graph of a random variable is always equal to 1. In the above graph, the area is: Watch the video for an overview and a few worked examples: Uniform Probability Distribution Examples Watch this video on YouTube. Can’t see the video? Click here. Note: In actuarial science, the uniform distribution is called the de Moivre distribution. TypesThis distribution has two types. The most common type you’ll find in elementary statistics is the continuous uniform distribution (in the shape of a rectangle). However, there is a second type: the discrete uniform distribution. It still resembles a rectangle but instead of a line, a series of dots represent a known, finite number of outcomes. The following graph shows 5 possible outcomes: Image: ikamusumeFan|Wikimedia CommonsRolling a single die is one example of a discrete uniform distribution; a die roll has four possible outcomes: 1,2,3,4,5, or 6. There is a 1/6 probability for each number being rolled. General FormulaThe general formula for the probability density function (pdf) for the uniform distribution is: Uniform CDFThe uniform distribution doesn’t always look like a rectangle. A special case, the uniform cumulative distribution function, adds up all of the probabilities (in the same way a cumulative frequency distribution adds probabilities) and plots the result, which is a linear graph and not a rectangle: VariablesThe variables in a uniform distribution are called uniform random variables. 2. Expected Value and Variance.The expected value (i.e. the mean) of a uniform random variable X is:
This is also written equivalently as: E(X) = (b + a) / 2. “a” in the formula is the minimum value in the distribution, and “b” is the maximum value. The variance of a uniform random variable is:
For the above image, the variance is (1/12)(3 – 1)2= 1/12 * 4 = 1/3. 3. Finding Probabilities for a Continuous Uniform DistributionNeed help with a homework question? Check out our tutoring page! Example question #1: The average amount of weight gained by a person over the winter months is uniformly distributed from 0 to 30lbs. Find the probability a person will gain between 10 and 15lbs during the winter months. Step 1: Find the height of the distribution. The area under a probability distribution is always 1. As there are 30 units (from zero to 30), then the height is 1/30. Step 2: Find the width of the “slice” of the distribution mentioned in the question. Do this by subtracting the biggest number (b) from the smallest (a), to get b – a = 15 – 10 = 5. Step 3: Multiply the width (Step 2) by the height (Step 1) to get: Example Question 2: Find P(X≤10) for the above question. Step 1: Find the width of the “box”: b – a = 10 – 0 = 10. Step 2: Multiply the width (Step 1) by the height. We already know the height is 1/30 (from example question 1), so: Example Question 3: Find P(20≤X≤25) for the above question. This is asking the probability of a weight gain between 20 and 25 pounds. Step 1: Find the width of the “box”: b – a = 25 – 20 = 5. Step 2: Multiply the width (Step 1) by the height. We already know the height is 1/30 (from example question 1), so: The More Formal FormulaYou can solve these types of problems using the steps above, or you can us the formula for finding the probability for a continuous uniform distribution:
This is also sometimes written as: ReferencesAgresti A. (1990) Categorical Data Analysis. John Wiley and Sons, New York.
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