University of California, Los AngelesDepartment of StatisticsStatistics 100B Instructor: Nicolas ChristouContinuous probability distributions• Let X be a continuous random variable, −∞ < X < ∞• f(x) is the so called probability density function (pdf) ifZ∞−∞f(x)dx = 1• Area under the pdf is equal to 1.• How do we compute probabilities? Let X be a continuous r.v.with pdf f(x). ThenP (X > a) =Z∞af(x)dxP (X < a) =Za−∞f(x)dxP (a < X < b) =Zbaf(x)dx• Note that in continuous r.v. the following is true:P (X ≥ a) = P (X > a)This is NOT true for discrete r.v.1• Cumulative distribution function (cdf):F (x) = P (X ≤ x) =Zx−∞f(x)dx• Thereforef(x) = F (x)0• Compute probabilities using cdf:P (a < X < b) = P (X ≤ b) − P (X ≤ a) = F (b) − F (a)• Example: Let the lifetime X of an electronic component inmonths be a continuous r.v. with f(x) =10x2, x > 10.a. Find P (X > 20).b. Find the cdf.c. Use the cdf to compute P (X > 20).d. Find the 75thpercentile of the distribution of X.e. Compute the probability that among 6 such electronic com-ponents, at least two will survive more than 15 months.2• Mean of a continuous r.v.µ = E(X) =Z∞−∞xf(x)dx• Mean of a function of a continuous r.v.E[g(X)] =Z∞−∞g(x)f (x)dx• Variance of continuous r.v.σ2= E(X − µ)2=Z∞−∞(x −µ)2f(x)dxOrσ2=Z∞−∞x2f(x)dx − [E(X)]2• Some properties: Let a, b constants and X, Y r.v.E(X + a) = a + E(X)E(X + Y ) = E(X) + E(Y )var(X + a) = var(X)var(aX + b) = a2var(X)If X, Y are independent thenvar(X + Y ) = var(X) + var(Y )3• Example: Let X be a continuous r.v. with f(x) = ax + bx2,and 0 < x < 1.a. If E(X) = 0.6 find a, b.b. Find var(X).4• Uniform probability distribution:A continuous r.v. X follows the uniform probability distributionon the interval a, b if its pdf function is given byf(x) =1b − a, a ≤ x ≤ b– Find cdf of the uniform distribution.– Find the mean of the uniform distribution.– Find the variance of the uniform distribution.5• The gamma distributionThe gamma distribution is useful in modeling skewed distribu-tions for variables that are not negative.A random variable X is said to have a gamma distribution withparameters α, β if its probability density function is given byf(x) =xα−1e−xββαΓ(α), α, β > 0, x ≥ 0.E(X) = αβ and σ2= αβ2.A brief note on the gamma function:The quantity Γ(α) is known as the gamma function and it isequal to:Γ(α) =Z∞0xα−1e−xdx.If α = 1, Γ(1) =R∞0e−xdx = 1.With integration by parts we get Γ(α + 1) = αΓ(α) as follows:Γ(α + 1) =Z∞0xαe−xdx =Let, v = xα⇒dvdx= αxα−1dudx= e−x⇒ u = −e−xTherefore,Γ(α + 1) =Z∞0xαe−xdx = −e−xxα|∞0−Z∞0−e−xαxα−1dx = αZ∞0xα−1e−xdx.Or, Γ(α + 1) = αΓ(a).6Similarly, using integration by parts it can be shown that,Γ(α + 2) = (α + 1)Γ(α + 1) = (α + 1)αΓ(α), and,Γ(α + 3) = (α + 2)(α + 1)αΓ(α).Therefore, using this result, when α is an integer we get Γ(α) =(α − 1)!.Example:Γ(5) = Γ(4 + 1) = 4 × Γ(4) = 4 × Γ(3 + 1) = 4 × 3 × Γ(3) =4×3×Γ(2+1) = 4×3×2×Γ(1+1) = 4×3×2××1×Γ(1) = 4!.Useful result:Γ(12) =√π.The gamma density for α = 1, 2, 3, 4 and β = 1.0 2 4 6 80.0 0.2 0.4 0.6 0.8 1.0xf(x)Gamma distribution densityΓΓ((αα == 1,, ββ == 1))ΓΓ((αα == 2,, ββ == 1))ΓΓ((αα == 3,, ββ == 1))ΓΓ((αα == 4,, ββ == 1))7• Exponential probability distribution:Useful for modeling the lifetime of electronic components.• A continuous r.v. X follows the exponential probability distri-bution with parameter λ > 0 if its pdf function is given byf(x) = λe−λx, x > 0Note: From the pdf of the gamma distribution, if we set α = 1and β =1λwe get f(x) = λe−λx. We see that the exponentialdistribution is a special case of the gamma distribution.– Find cdf of the exponential distribution.– Find the mean of the exponential distribution.– Find the variance of the exponential distribution.– Find the median of the exponential distribution.– Find the pthpercentile of the exponential distribution.8• Example:Let X be an exponential random variable with λ = 0.2.a. Find the mean of X.b. Find the median of X.c. Find the variance of X.d. Find the 80thpercentile of this distribution (or find c suchthat P (X < c) = 0.80).9• Memoryless property of the exponential distribution:Suppose the lifetime of an electronic component follows the expo-nential distribution with parameter λ. The memoryless propertystates thatP (X > s + t|X > t) = P (X > s), s > 0, t > 0Example:Suppose the number of miles a car can run before its battery wears out follows theexponential distribution with mean µ = 10000 miles. If the owner of the car takes a5000-mile trip what is the probability that he will be able to complete the trip withouthaving to replace the battery of the car?If the number of miles follow some other distribution with known cumulative distribu-tion function (cdf) give an expression of the probability of completing the trip withouthaving to replace the battery of the car in terms of the cdf.10The distribution of a function of a random variablesSuppose we know the pdf of a random variable X. Many times we want to find the proba-bility density function (pdf) of a function of the random variable X. Suppose Y = Xn.We begin with the cumulative distribution function of Y :FY(y) = P (Y ≤ y) = P (Xn≤ y) = P (X ≤ y1n).So far we haveFY(y) = FX(y1n)To find the pdf of Y we simply differentiate both sides wrt to y:fY(y) =1ny1n−1× fX(y1n).where, fX(·) is the pdf of X which is given. Here are some more examples.Example 1Suppose X follows the exponential distribution with λ = 1. If Y =√X find the pdf of Y .Example 2Let X ∼ N(0, 1). If Y = eXfind the pdf of Y . Note: Y it is said to have a log-normaldistribution.Example 3Let X be a continuous random variable with pdf f(x) = 2(1 −x), 0 ≤ x ≤ 1. If Y = 2X −1find the pdf of Y .Example 4Let X be a continuous random variable with pdf f (x) =32x2, −1 ≤ x ≤ 1. If Y = X2findthe pdf of Y .11Continuous random variables - Some examples(Some are from: Sheldon Ross (2002), A first Course in Probability, Sixth Edition, Prentice Hall).Example 1Suppose X, the lifetime of a certain type of electronic device (in hours), is a continuous random variablewith probability density function f(x) =10x2for x > 10 and f (x) = 0 for x ≤ 10.a. Find P (X > 20).b. Find the cumulative distribution function