Random Variables, Expected Values and Variance von Edu Pristine

Dozent des Vortrages Random Variables, Expected Values and Variance

Edu Pristine

Trusted by Fortune 500 Companies and 10,000 Students from 40+ countries across the globe, EduPristine is one of the leading International Training providers for Finance Certifications like FRM®, CFA®, PRM®, Business Analytics, HR Analytics, Financial Modeling, Operational Risk Modeling etc. It was founded by industry professionals who have worked in the area of investment banking and private equity in organizations such as Goldman Sachs, Crisil - A Standard & Poors Company, Standard Chartered and Accenture.

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... distribution, the Bernoulli distribution, the binomial distribution, the geometric distribution, and the negative binomial distribution all uses discrete random variable. Examples: The number of brown eyed children in a family containing two children is a discretevariable as its set of possible values is a finite, countable set. The set contains the values, {0,1,2} , that is either in a family none of the child has a brown eye, either one of the child has a brown eye or both children have a brown eye ...

... The variable is said to be a continuous random variable if it can assume any value insome interval. Examples: distance, speed, time, etc.The probability it takes at any point is assumed to be in case of continuous Random Variable The normal distribution, continuous uniform distribution, Beta distribution, and Gamm a distribution are all continuous distributions. Useful when variable holds value between two limits. P[a

... number of jacks? In the above example the number of jacks can be taken as a random variable, X. Clearly X can take values 0 (no jack out of two cards drawn), 1(one jack), 2 ( both cards draw n are jack) P(X=0) = P(non-jack and non-jack) = P(non-jack)*P(non-jack) = 48/52 * 48 /52 = 144/169 P(X=1) = P(jack and non jack or non-jack or jack) = P(jack and non-jack) + P(non-jack and jack) = 4/52*48 /52 + 48/52*4/52 = 24/169 P(X=2) = 4/52 * 4/52 = 1/169 ...

... continuous variable found using integration over a limit at a particular value the probability measured is zero Hence P(x1 "d X "d x2) = P(x1 < X "d x2 ) = P( x1 "d X < x2) " Example: The PDF of a continuous variable is given by 6x 5 and 0 "d x "d 1 What is the probability that 0 < X < 0.5, the probability that X lies between 0 and 0.5. Cumulative probability distribution. A cumulative probability density function for a r.v. is defined ...

... distribution are stable, then the distribution itsel f is said to be a stable distribution. If X 1and X 2are the independent random variables from a stable distribution, and a, b are scalar variables, then the variable Y = a.X 1+ b.X 2also follows the same distribution. The Expected value of Y: E(Y) = a.E(X1) + b.E(X 2) " The Variance of Y: Var(Y)= a2 .Var(X 1) + b 2 .Var(X 2) ...

... Knowledge Management Pristine7 Scalar multiplication of a random number. On multiplication of a discrete random variable by a fixed number, the probabilities remain unchanged but the possibilities are multiplied by that fixed number. Example: Suppose X is a discrete random variable, then 2X is defined by the same probabilitydistribution as , that ...

... probabilit xof value. Expected value E[X] = If the probabilities of obtaining the amounts a 1, a 2, a 3, &,a k, are p 1, p 2, p 3, &,p k. Then the expected value is E = a 1* p 1+ a 2* p 2+ a 3* p 3+ &.+ a k* p k. Example: A drug is used to maintain a steady rate in patients who have suffered a mil d heart attack. Let X denote the number of heartbeats per minute obtained per ...

... X = ? 2 = E[X 2 ] (E[X]) 2 Standard Deviatio. The positive root of variance Not all random variables have a standard deviation, since these expected values need not exist If the variable is continuous, in that case we replace summation with integration Example: Find the variance and standard deviation for previous example E[X] = 35, E[X 2 ] = 1231.6 Variance = 1231.6 35*35 = ..