What are the conditions of probability distribution?
Learning Objectives Associated to each possible value \(x\) of a discrete random variable \(X\) is the probability \(P(x)\) that \(X\) will take the value \(x\) in one trial of the experiment. Definition: probability distribution The probability distribution of a discrete random variable \(X\) is a list of each possible value of \(X\) together with the probability that \(X\) takes that value in one trial of the experiment. The probabilities in the probability distribution of a random variable \(X\) must satisfy the following two conditions:
Example \(\PageIndex{1}\): two Fair Coins A fair coin is tossed twice. Let \(X\) be the number of heads that are observed.
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Example \(\PageIndex{2}\): Two Fair Dice A pair of fair dice is rolled. Let \(X\) denote the sum of the number of dots on the top faces.
Solution: The sample space of equally likely outcomes is \[\begin{matrix} 11 & 12 & 13 & 14 & 15 & 16\\ 21 & 22 & 23 & 24 & 25 & 26\\ 31 & 32 & 33 & 34 & 35 & 36\\ 41 & 42 & 43 & 44 & 45 & 46\\ 51 & 52 & 53 & 54 & 55 & 56\\ 61 & 62 & 63 & 64 & 65 & 66 \end{matrix} \nonumber\] where the first digit is die 1 and the second number is die 2.
The Mean and Standard Deviation of a Discrete Random VariableDefinition: mean The mean (also called the "expectation value" or "expected value") of a discrete random variable \(X\) is the number \[\mu =E(X)=\sum x P(x) \label{mean}\] The mean of a random variable may be interpreted as the average of the values assumed by the random variable in repeated trials of the experiment. Example \(\PageIndex{3}\) Find the mean of the discrete random variable \(X\) whose probability distribution is \[\begin{array}{c|cccc} x &-2 &1 &2 &3.5\\ \hline P(x) &0.21 &0.34 &0.24 &0.21\\ \end{array} \nonumber\] Solution Using the definition of mean (Equation \ref{mean}) gives \[\begin{align*} \mu &= \sum x P(x)\\[5pt] &= (-2)(0.21)+(1)(0.34)+(2)(0.24)+(3.5)(0.21)\\[5pt] &= 1.135 \end{align*}\] Example \(\PageIndex{4}\) A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for \(\$1\) each. Each has an equal chance of winning. First prize is \(\$300\), second prize is \(\$200\), and third prize is \(\$100\). Let \(X\) denote the net gain from the purchase of one ticket.
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The concept of expected value is also basic to the insurance industry, as the following simplified example illustrates. Example \(\PageIndex{5}\) A life insurance company will sell a \(\$200,000\) one-year term life insurance policy to an individual in a particular risk group for a premium of \(\$195\). Find the expected value to the company of a single policy if a person in this risk group has a \(99.97\%\) chance of surviving one year. Solution: Let \(X\) denote the net gain to the company from the sale of one such policy. There are two possibilities: the insured person lives the whole year or the insured person dies before the year is up. Applying the “income minus outgo” principle, in the former case the value of \(X\) is \(195-0\); in the latter case it is \(195-200,000=-199,805\). Since the probability in the first case is 0.9997 and in the second case is \(1-0.9997=0.0003\), the probability distribution for \(X\) is: \[\begin{array}{c|cc} x &195 &-199,805 \\ \hline P(x) &0.9997 &0.0003 \\ \end{array}\nonumber \] Therefore \[\begin{align*} E(X) &=\sum x P(x) \\[5pt]&=(195)\cdot (0.9997)+(-199,805)\cdot (0.0003) \\[5pt] &=135 \end{align*}\] Occasionally (in fact, \(3\) times in \(10,000\)) the company loses a large amount of money on a policy, but typically it gains \(\$195\), which by our computation of \(E(X)\) works out to a net gain of \(\$135\) per policy sold, on average. Definition: variance The variance (\(\sigma ^2\)) of a discrete random variable \(X\) is the number \[\sigma ^2=\sum (x-\mu )^2P(x) \label{var1}\] which by algebra is equivalent to the formula \[\sigma ^2=\left [ \sum x^2 P(x)\right ]-\mu ^2 \label{var2}\] Definition: standard deviation The standard deviation, \(\sigma \), of a discrete random variable \(X\) is the square root of its variance, hence is given by the formulas \[\sigma =\sqrt{\sum (x-\mu )^2P(x)}=\sqrt{\left [ \sum x^2 P(x)\right ]-\mu ^2} \label{std}\] The variance and standard deviation of a discrete random variable \(X\) may be interpreted as measures of the variability of the values assumed by the random variable in repeated trials of the experiment. The units on the standard deviation match those of \(X\). Example \(\PageIndex{6}\) A discrete random variable \(X\) has the following probability distribution: \[\begin{array}{c|cccc} x &-1 &0 &1 &4\\ \hline P(x) &0.2 &0.5 &a &0.1\\ \end{array} \label{Ex61}\] A histogram that graphically illustrates the probability distribution is given in Figure \(\PageIndex{3}\). Figure \(\PageIndex{3}\): Probability Distribution of a Discrete Random VariableCompute each of the following quantities.
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Summary
What are the three required conditions of a probability distribution?The requirements for a probability distribution are: 1) The value of the random variable X must be described. 2) All the values of random variable X must have their corresponding probabilities. 3) The sum of all probabilities is equal to one.
What are the properties of a probability distribution?A probability distribution depicts the expected outcomes of possible values for a given data generating process. Probability distributions come in many shapes with different characteristics, as defined by the mean, standard deviation, skewness, and kurtosis.
What is meant by conditional probability distribution?Informally, we can think of a conditional probability distribution as a probability distribution for a sub-population. In other words, a conditional probability distribution describes the probability that a randomly selected person from a sub-population has a given characteristic of interest.
How do you know if its a probability distribution or not?Step 1: Determine whether each probability is greater than or equal to 0 and less than or equal to 1. Step 2: Determine whether the sum of all of the probabilities equals 1. Step 3: If Steps 1 and 2 are both true, then the probability distribution is valid. Otherwise, the probability distribution is not valid.
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