2.6. Probability. In general the probability density of a random variable satisfies two conditions:


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1 2.6. PROBABILITY Probability Continuous Random Variables. A random variable a realvalued function defined on some set of possible outcomes of a random experiment; e.g. the number of points obtained after rolling a dice  which can be, 2, 3, 4, 5 or 6. In th example the random variable can take only a dcrete set of values. If the variable can take a continuous set of values then it called a continuous random variable, e.g. a person s height. Given a random variable X, its probability dtribution function the function F (x) = P (X x) = probability that the random variable X takes a value less than or equal to x. For instance if F (x) the probability dtribution function of the number of points obtained after rolling a dice, then F (4.7) = P (X 4.7) = probability that the number of points less than or equal to 4.7, i.e., the number of points, 2, 3 or 4, so the probability 4/6 = 2/3, and F (4.7) = 2/3. If the random variable continuous then we can also define a probability density function f(x) equal to the limit as x of the probability that the random variable takes a value in a small interval of length x around x divided by the length of the interval. Th definition means that f(x) = F (x). The probability that the random variable takes a value in some interval [a, b] P (a X b) = F (b) F (a) = b a f(x) dx. In general the probability density of a random variable satfies two conditions: () f(x) for every x (probabilities are always nonnegative). (2) f(x) dx = (the probability of a sure event ). Example: A probability dtribution called uniform on a set S if its probability density constant on S. Find the probability density of the uniform dtribution on the interval [2, 5]. Answer: The probability density function must be constant on [2, 5], so for 2 x 5 we have f(x) = c for some constant c. On the other
2 hand f(x) = for x outside [2, 5], hence: = 2.6. PROBABILITY 67 f(x) dx = 5 2 c dx = c(5 2) = 3c, so c = /3. Hence f(x) = /3 if 2 x 5, otherwe Means. The mean or average of a dcrete random variable that takes values x, x 2,..., x n with probabilities p, p 2,..., p n respectively n x = x p + x 2 p x n p n = x i p i. For instance the mean value of the points obtained by rolling a dice i= = 7 2 = 3.5. Th means that if we roll the dice many times in average we may expect to get about 3.5 points per roll. For continuous random variables the probability replaced with the probability density function, and the sum becomes an integral: µ = x = xf(x) dx Waiting Times. The time that we must wait for some event to occur (such as receiving a telephone call) can be modeled with a random variable of density if t <, f(t) = ce ct if t, were c a positive constant. Note that, as expected: f(t) dt = ce ct dx = [ e ct] = lim u e cu ( e )} =.
3 2.6. PROBABILITY 68 The mean waiting time can be computed like th: µ = = tf(t) dt tce ct dx = [ te ct] + e ct dx = + = c, ] [ e ct c (I. by parts) hence µ = /c. So we can rewrite the density function like th: if t <, f(x) = µ e t/µ if t, Example: Assume that the average waiting time for a catastrophic meteorite to strike the Earth million years. Find the probability that the Earth will suffer a catastrophic meteorite impact in the next years. Find the probability that no such catastrophic event will happen in the next 5 billion years. Answer: We have µ = 8 years, so the probability density function f(t) = 8 e 8 t So the answer to the first question 8 e 8 t dt = [ ] e 8 t (t ). = e 8 = e 6 6, i.e., about in a million. Regarding the second question, the probability 5 9 [ ] e 8 t dt = e 8 t = e 8 5 9} = e , which practically zero.
4 2.6. PROBABILITY Normal Dtributions. Many important phenomena follow a so called Normal Dtribution, whose density function : f(x) = σ /2σ2 e (x µ)2. 2π Its mean µ. The positive constant σ called standard deviation; it measures how spread out the values of the random variable are. Example: Intelligent Quotient (IQ) scores are dtributed normally with mean µ = and standard deviation σ = 5. What proportion of the population has an IQ between 7 and 3? Answer: The integral cannot be evaluated in terms of elementary functions, but it can be approximated with numerical methods: P (7 X 3) = π e (x )2 /2 52 dx Another approach for solving these kinds of problems to use the error function, defined in the following way: φ(x) = 2 x e t2 dt. π The following are some of its values (rounded to three decimal places): x φ(x) x φ(x) x φ(x) Example: Solve the previous problem using the error function. Answer: We need to transform our integral into another expression containing the error function:
5 2.6. PROBABILITY 7 P (7 X 3) = 3 7 = π 2 5 2π e (x )2 /45 dx 2 = 2 2 e u2 du π e u2 du [u = (x )/5 2] = φ( 2) φ(.4).952 (by symmetry)
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