University of Utah



Final – Practice ExamQuestion #1: If X1 and X2 denotes a random sample of size n=2 from the population with density function fXx=1x21{x>1}, then a) find the joint density of U=X1X2 and V=X2, and b) find the marginal probability density function of U.By the independence of the two random variables X1 and X2, we have that their joint density is given by fX1X2x1,x2=fX1x1fX2x2=x1-2x2-2=x1x2-2 whenever x1>0 and x2>0. Then u=x1x2 and v=x2 implies that x2=v and x1=ux2=uv, so we can compute the Jacobian as J=det?x1?u?x1?v?x2?u?x2?v=det1v-uv201=1v. This implies that the joint density of U and V is fUVu,v=fX1X2uv,vJ=uvv-21v=1u2v whenever v>1 and uv>1→u>v. This also shows U and V are independent.We have fUu=1ufUVu,vdv=1u1u2vdv=1u2ln(v)1u=lnuu2 if u>1.Question #2: Find the Maximum Likelihood Estimators (MLE) of θ and α based on a random sample X1,…,Xn from fXx;θ,α=αθαxα-11{0≤x≤θ} where θ>0 and α>0.We first note that the likelihood function is given by Lθ,α=i=1nfXxi;θ,α=i=1nαθ-αxiα-11{0≤xi≤θ}=αnθ-nαi=1nxiα-110≤x1:n1xn:n≤θ. Then by inspection, we see that θMLE=Xn:n. To find the MLE of α, we compute lnLθ,α=nln(α)-nαlnθ+(α-1)i=1nln(xi) so that the derivative with respect to α is ??αlnLθ,α=nα-nlnθ+i=1nlnxi=0→α=nnlnθ-i=1nlnxi. Since the second derivative is zero, we have found that αMLE=nnlnXn:n-i=1nlnXi.Question #3: If X1,…,Xn is a random sample from N(μ,σ2), then a) find a UMVUE for μ when σ2 is known, and b) find a UMVUE for σ2 when μ is known.We have previously shown that the Normal distribution is a member of the Regular Exponential Class (REC), where S1=i=1nXi and S2=i=1nXi2 are jointly complete sufficient statistics for the unknown parameters μ and σ2. Since we have that ES1=Ei=1nXi=i=1nE(Xi)=i=1nμ=nμ, the statistic T=1nS1=1ni=1nXi=X will be unbiased for the unknown parameter μ. The Lehmann-Scheffe Theorem then guarantees that T=X is a UMVUE for μ when σ2 is known.Since ES2=Ei=1nXi2=i=1nE(Xi2)=i=1nVarXi+EXi2=i=1nσ2+μ2=nσ2+μ2, the statistic T=1nS2-μ2 will be unbiased for the unknown parameter σ2 when μ is known. The Lehmann-Scheffe Theorem states that T is a UMVUE for σ2.Question #4: Consider a random sample X1,…,Xn from fXx;θ,α=αθαxα-11{0≤x≤θ}, where θ>0 is unknown but α>0 is known. Find the constants 1<a<b depending on the values of n and α such that (aXn:n,bXn:n) is a 90% confidence interval for θ.We first compute the CDF of the population FXx;θ,α=0xfX(t;θ,α)dt=αθα0xxα-1dt=αθα1αxα0x=xθα whenever 0≤x≤θ. Then we use this to compute the CDF of the largest order statistic FXn:nx=PXn:n≤x=PXi≤xn=xθnα. In order for (aXn:n,bXn:n) to be a 90% confidence interval for θ, it must be the case that PaXn:n≥θ=0.05 and PbXn:n≤θ=0.05. We solve each of these two equations for the unknown constants, so PaXn:n≥θ=0.05→PXn:n≥θa=0.05→1-PXn:n≤θa=0.95→1-FXn:nθa=0.95→1-θ/aθnα=0.95→1-1anα=0.95→a=10.951nα=20191nα . Similarly, we find that b=201nα.Question #5: If X1 and X2 are independent and identically distributed from EXP(θ) such that their density is fXx=1θe-xθ1{x>0}, then find the joint density of U=X1 and V=X2X1.We have fUVu,v=fX1X2u,uvdet10vu=1θe-uθ1θe-uvθu=uθ2e-u(1-v)θ whenever we have that u>0 and uv>0→v>0. Question #6: Let X1,…,Xn be a random sample from WEI(θ,β) with known β such that their common density is fXx;θ,β=βθβxβ-1e-xθβ1{x>0}. Find the MLE of θ.We have Lθ,β=i=1nβθβxiβ-1e-xiθβ=βθ-βni=1nxiβ-1e-1θβi=1nxiβ so that lnLθ,β=nlnβ-nβlnθ+β-1i=1nlnxi-1θβi=1nxiβ. Then we can calculate that the ??θlnLθ,β=-nβθ+βθβ+1i=1nxiβ=0→nβθ=βθβ+1i=1nxiβ→n=1θβi=1nxiβ→θβ=i=1nxiβn. Therefore, we have found θMLE=1ni=1nXiβ1β.Question #7: Use the Cramer-Rao Lower Bound to show that X is a UMVUE for θ based on a random sample of size n from an EXP(θ) distribution where fXx;θ=1θe-xθ1{x>0}.We first find the Cramer-Rao Lower Bound; since τθ=θ, we have τ'(θ)2=1. Then we have lnfX;θ=-lnθ-Xθ so that ??θlnfX;θ=-1θ+Xθ2 and ?2?θ2lnfX;θ=1θ2-2Xθ3. Finally, -E?2?θ2lnfX;θ=-1θ2+2θ3EX=2θθ3-1θ2=1θ2 implies that CRLB=θ2n. To verify that X is a UMVUE for θ, we first show that EX=E1ni=1nXi=1ni=1nE(Xi)=1nnθ=θ. Then we check that the variance achieves the Cramer-Rao Lower Bound from above, so VarX=Var1ni=1nXi=1n2i=1nVar(Xi)=1n2nθ2=θ2n=CRLB. Thus, X is a UMVUE for θ.Question #8: Use the Lehmann-Scheffe Theorem to show that X is a UMVUE for θ based on a random sample of size n from an EXP(θ) distribution where fXx;θ=1θe-xθ1{x>0}.We know that EXP(θ) is a member of the Regular Exponential Class (REC) with t1x=x, so the statistic S=i=1nt1(Xi)=i=1nXi is complete sufficient for the parameter θ. Then ES=Ei=1nXi=i=1nE(Xi)=nθ implies that the estimator T=1nS=1ni=1nXi=X is unbiased for θ, so the Lehmann-Scheffe Theorem guarantees that it will be the Uniform Minimum Variance Unbiased Estimator of θ.Question #9: Find a 1001-α% confidence interval for θ based on a random sample of size n from an EXP(θ) distribution where fXx=1θe-xθ1{x>0}. Use the facts that EXPθ≡GAMMA(θ,1), that θ is a scale parameter, and that GAMMA2,n≡χ2(2n).Since θ is a scale parameter, we know that Q=θMLEθ=Xθ=i=1nXinθ will be a pivotal quantity. Note that we used the fact that θMLE=X, which was derived in a previous exercise. We begin by noting that since each Xi~EXPθ≡GAMMA(θ,1), we can conclude that A=i=1nXi~GAMMA(θ,n). In order to obtain a chi-square distributed pivotal quantity, we use the modified pivot R=2nQ=2θi=1nXi. To verify its distribution, we use the CDF technique so FRr=PR≤r=P2θi=1nXi≤r=Pi=1nXi≤θr2=FAθr2 implies that fRr=ddrFAθr2=fAθr2θ2=1θnΓnθr2n-1e-θr2θθ2=12nΓ(n)rn-1e-r2. This proves that R=2θi=1nXi=2nXθ~χ2(2n) so Pχα222n<2nXθ<χ1-α222n=1-α→P2nXχ1-α222n<θ<2nXχα222n=1-α is the desired 1001-α% confidence interval for the unknown parameter θ.Question #10: Let X1,…,Xn be independent and identically distributed from the population with CDF given by FXx=11+e-x. Find the limiting distribution of Yn=X1:n+ln(n).We have that FYny=PYn≤y=PX1:n+lnn≤y=PX1:n≤y-lnn=1-PX1:n>y-lnn=1-PXi>y-lnnn=1-1-FXy-lnnn=1-1-11+e-y+lnnn=1-elnn-y1+elnn-yn=1-ne-y1+ne-yn=1-1+1ne-yn=1-1+1ne-yn=1-1+eynn. Then limn→∞FYny=1-limn→∞1+eynn=1-eey is the limiting distribution since limn→∞1+cnnb=ecb for all real numbers c and b.Question #11: Let X1,…,Xn be independent and identically distributed from the population with a PDF given by fXx=121{-1<x<1}. Approximate P(i=1100Xi≤0) using Φ(z).From the given density, we know that X~UNIF(-1,1) so EX=0 and VarX=13. Then if we define Y=i=1100Xi, we have that EY=Ei=1100Xi=i=1100E(Xi)=0 and VarY=Vari=1100Xi=i=1100Var(Xi)=100/3. These facts allow us to compute Pi=1100Xi≤0=PY≤0=PY-0100/3≤0-0100/3≈PZ≤0=Φ0=1/2.Question #12: Suppose that X is a random variable with density fXx=12e-x for x∈R. Compute the Moment Generating Function (MGF) of X and use it to find E(X2).By definition, we have MXt=EetX=-∞∞etxfX(x)dx=-∞∞etx12e-x dx=120∞etxe-xdx+12-∞0etxexdx=120∞ex(t-1)dx+12-∞0ex(t+1)dx. After integrating and collecting like terms, we obtain MXt=12t+1-1-t-1-1. We then compute ddtMXt=12-t+1-2+t-1-2 and d2dt2MXt=122t+13-2t-13 so that EX2=d2dt2MX0=1220+13-20-13=122+2=2.Question #13: Let X be a random variable with density fXx=141{-2<x<2}. Compute the probability density function of the transformed random variable Y=X3.We have FYy=PY≤y=PX3≤y=PX≤y13=FXy13 so that fYy=ddyFXy13=fXy1313y-23=1413y-23=112y23 if -2<y13<2→-8<y<8.Question #14: Let X1,…,Xn be a random sample from the density fXx=θxθ-1 whenever 0<x<1 and zero otherwise. Find a complete sufficient statistic for the parameter θ.We begin by verifying that the density is a member of the Regular Exponential Class (REC) by showing that fXx=explnθxθ-1=exp{lnθ+θ-1ln(x)}=θexp{θ-1ln(x)}, where q1θ=θ-1 and t1x=ln(x). Then we know that the statistic S=i=1nt1(Xi)=i=1nln(Xi) is complete sufficient for the parameter θ.Question #15: Let X1,…,Xn be a random sample from the density fXx=θxθ-1 whenever 0<x<1 and zero otherwise. Find the Maximum Likelihood Estimator (MLE) for θ>0.We have Lθ=i=1nfX(xi)=i=1nθxiθ-1=θni=1nxiθ-1 so that lnLθ=nln(θ)+(θ-1)i=1nln(xi) and ??θlnLθ=nθ+i=1nln(xi)=0→θ=-ni=1nln(xi). Thus, we have found that the Maximum Likelihood Estimator is θMLE=-ni=1nln(Xi).Question #16: Let X1,…,Xn be a random sample from the density fXx=θxθ-1 whenever 0<x<1 and zero otherwise. Find the Method of Moments Estimator (MME) for θ>0.We have EX=01xfXxdx=01xθxθ-1dx=θ01xθdx=θθ+1xθ+101=θθ+1, so that μ1'=M1'→θθ+1=X→θ1-X=X so the desired estimator is θMME=X1-X. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download