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UCLA STATS 100A - BEAMER-UNIFORM

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Continuous Random Variables: The Uniform.Juana [email protected] Department of StatisticsJ. Sanchez Continuous Random VariablesUniform family of continuous random variablesI. The Uniform random variable.J. Sanchez Continuous Random VariablesI. The Uniform random variableExperiments that consist of observing events in a certain time frame,such as buses arriving at a bus stop or telephone calls coming into aswitchboard during a specified period, are sometimes modeled with aUniform r.v. Suppose that we know that one such event has occurred inthe time interval α, β : a bus arrived between 8 and 8:10. It may then beof interest to place a probability density on the actual time of occurrenceof the event under observation, which would be the r.v. X. The Uniformmodel assumes that X is equally likely to lie in any small subinterval.J. Sanchez Continuous Random VariablesPdf and CDF of a Uniform random variableA uniform random variable X with parameters α and β can becharacterized as follows:Density function (pdf)f (x) =1β−αα < x < β0 otherwiseCumulative Distribution Function (CDF)F (x ) = Prob(X ≤ x ) =Rxα1β−αdx=x −αβ−αα < x < β,= 0 otherwiseand F0(x ) =1β−α= f (x)We can use the cdf to compute probabilities of uniform random variablesP(a < X < b) = F (b) − F (a)J. Sanchez Continuous Random VariablesFigure : Uniform pdf (α = 1, β = 4)Figure : CDF of auniform(α = 1, β = 4)J. Sanchez Continuous Random VariablesFinding probabilities for Uniform random variablesProbabilitiesP(a < X < b) =Zba1β − αdx = (b − a)1β − αIf a = α, b = β then the integral is 1.You may play with several values of a and b. But notice also that we canuse the cumulative distribution function to compute probabilitiesP(a < X < b) = F (b) − F (a)J. Sanchez Continuous Random VariablesExampleLet X be a uniformly distributed random variable over (0,10). Calculatethe following probabilities(a)P(X < 3) =Z30110dx = 3/10(b)P(X > 6) =Z106110dx = 4/10(c)P(3 < X < 8) =Z83110dx = 1/2J. Sanchez Continuous Random VariablesExampleA bomb is to be dropped along a 1-mile-long line that stretches across apractice target zone. The target zones center is at the midpoint of theline. The target will be destroyed if the bomb falls within 1/10 mile oneither side of the center. Find the probability that the target will bedestroyed, given that the bomb falls at a random location along the line.X ∼ U (−0.5, 0.5) reads as ” let Xbe Uniform in the interval−0.5, 0.5. Thusf (x) = 1 − 0.5 < X < 0.5P(−0.1 < X < 0.1) =R0.1−0.11dx = x |0.1−0.1= 0.2Alternatively,P(−0.1 < X < 0.1) =F (0.1) − F (−0.1) =0.1 + 0.51−−0.1 + 0.51= 0.2Figure : P (−0.1 < X < 0.1)J. Sanchez Continuous Random VariablesNotice that we could have done the last problem by assuming that thedistribution goes from 0 to 1. In that case, the center is at 0.5 and wewould compute the probability that X is between 0.4 and 0.6J. Sanchez Continuous Random VariablesExampleThe amount of time, in hours, that a computer functions before breakingdown is a continuous r.v. with pdff (x) =11000 ≥ x ≤ 100This is the uniform random variable.What is the cumulative distribution function for this continuous r.v.?F (x ) =Zx01100dt =x100Check that we did the right thing by taking the derivative of yourcumulative distribution function and seeing whether it is equal tof(x).Compute P(4 ≤ X ≤ 10) using the cumulative distribution function.Interpret what your computation means in the context of thisproblem.P(4 ≤ X ≤ 10) = F (10) − F (4) =10100−4100=6100= 0.06Probability that it breaks down after 4 hours of use but no later than6 hours is small, 0.06.J. Sanchez Continuous Random VariablesMoment Generating FunctionMgfM(t ) = E (etX)=ZβαetX1β − αdx=1β − αZβαetXdx=1β − αetXt|βα=etβ− etαt(β − α)J. Sanchez Continuous Random VariablesExpected Value of a uniform random variable using thedefinition of expected valueLet X ∼ U (α, β), thenµx=Z∞−∞x1β − αdx=Zβαx1β − αdx=β2− α22(β − α)=β + α2Notice how the expected value of a uniform r.v. is the mid point of theinterval in which the uniform random variable is defined.J. Sanchez Continuous Random VariablesVariance of uniform random variablesσ2x= E (x2) − [µx]2E (X2) =Zβαx21β − αdx =β3− α33(β − α)=β2+ αβ + α23σ2xσ2x=β2+ αβ + α23−(α + β)24=(β − α)212J. Sanchez Continuous Random VariablesUsing the Moment generating function to deriveMoments of the uniform r.v.We can also derive the expected value and variance of a uniform randomvariable from the moment generating function.µx= M0(t) |t=0=(b − a)t(betb− aeta) − (etb− eta)(b − a)(b − a)2t2|t=0=t(betb− aeta) − (etb− eta)(b − a)t2|t=0Because this is undefined, take the derivative of the numerator anddenominator, and evaluate at t=0 (L’hospital rule). First, takederivatives:N0D0=(betb− aeta) + t(b2etb− a2eta) − (betb− aeta)2t(b − a)=b2etb− a2eta2(b − a)Evaluate that expression at t=0:b2− a22(b − a)=b + a2J. Sanchez Continuous Random VariablesPercentiles of a uniform random variableThe amount of time, in hours, that a computer functions before breakingdown is a continuous r.v. with pdff (x) =11000 < X < 100Find the 90th percentile and the Interquartile Range.(a) 90th percentileRc01100dx = 0.9. Solve for c .c − 0100 − 0= 0.9 → c = 90(b) 25th percentile or lower quartilec − 0100 − 0= 0.25 → c = 25(c) 75th percentile or third quartilec − 0100 − 0= 0.75 → c = 75So IQR= 75-25=50J. Sanchez Continuous Random VariablesThe special case of the Uniform(0,1)A random variable is said to be uniformly distributed over the interval(0,1) if its pdf is given byf (x) =1 0 < x < 10 otherwiseLet (a, b) be a subinterval of (0,1). ThenP(a < X < b) =Zba1dx = b − aJ. Sanchez Continuous Random


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UCLA STATS 100A - BEAMER-UNIFORM

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