Question

Hello, can you please solve this problem? Thank you!

3.7 Find the mean and variance of the random variable x for the following cases: (a) x is a uniformly distributed random variable, whose pdf is 2 (P3.3) otherwise Also consider the special case when a =-b. (b) x is a Rayleigh distributed random variable, whose pdf is 'x > 0 (P3.4) 0 otherwise (c) x is a Laplacian distributed random variable, whose pdf is (P3.5) 2 (d) y is a discrete random variable, given by y-xi, where the random vari ables xi, 1-1, 2, . . . ,n, are statistically independent and identically distributed (1.1.d.) random variables with the pmf: P(x; 1) = p and P(xi = 0) = 1-p Remark y is, in fact, a binomial distributed random variable. However, you do not need the binomial distribution to calculate the mean and variance of y. All you need are the linear property of the expectation operation Es-} and the fact that if U and V are statistically independent random variables, then EtUV] -

Answer 1

$(a)~E(X)=\frac{1}{b-a}\int_a^bx dx=\frac{1}{b-a}\times \frac{b^2-a^2}{2}=\frac{a+b}{2}\\ E(X^2)=\frac{1}{b-a}\int_a^bx^2 dx=\frac{1}{b-a}\times \frac{b^3-a^3}{3}=\frac{a^2+b^2+ab}{3}\\ Var(X)=E(X^2)-E^2(X)=\frac{a^2+b^2+ab}{3}-\left(\frac{a+b}{2} \right )^2\\ =\frac{4a^2+4b^2+4ab-3a^2-3b^2-6ab}{12}=\frac{(b-a)^2}{12}\\ If~a=-b~then~E(X)=0,~Var(X)=\frac{b^2}{3}\\ (b)~E(X)=\int_0^\infty \frac{x^2}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}dx\\ =\sqrt{2}\sigma\int_0^\infty t^{3/2-1} e^{-t}dt,~~take,~\frac{x^2}{2\sigma^2}=t~or,~\frac{x}{\sigma^2}dx=dt,~x=\sqrt{2}\sigma \sqrt{t}\\ =\sqrt{2}\sigma\Gamma(3/2)=\sqrt{2}\sigma\times (1/2)\times \sqrt{\pi}=\sqrt{\frac{\pi}{2}}\sigma\\ E(X^2)=\int_0^\infty \frac{x^3}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}dx\\ =2\sigma^2\int_0^\infty t^{2-1} e^{-t}dt,~~take,~\frac{x^2}{2\sigma^2}=t~or,~\frac{x}{\sigma^2}dx=dt,~x=\sqrt{2}\sigma \sqrt{t}\\ =2\sigma^2\sigma\Gamma(2)=2\sigma^2\\ Var(X)=E(X^2)-E^2(X)=2\sigma^2-\frac{\pi}{2}\sigma^2=\frac{4-\pi}{2}\sigma^2\\$

$(c)~E(X)=\frac{c}{2}\int_{-\infty}^\infty xe^{-c|x|}dx=0~since~xe^{-c|x|}~is~odd~function~in~x.\\ Var(X)=E(X^2)=\frac{c}{2}\int_{-\infty}^\infty x^2e^{-c|x|}dx=c\int_{0}^\infty x^2e^{-cx}dx~since~x^2e^{-c|x|}~is~even~function~in~x.\\ =c\frac{\Gamma(3)}{c^3}=\frac{2}{c^2}\\ (d)~E(y)=E\left(\sum_{i=1}^n y_i \right )=\sum_{i=1}^nE(x_i)=\sum_{i=1}^nP(x_i=1)=np\\ Var(y)=Var\left(\sum_{i=1}^n y_i \right )=\sum_{i=1}^nVar(x_i);~since~x_i's~are~independent\\ =np(1-p)\\ ~since~Var(x_i)=E(x_i^2)-E^2(x_i)\\ =1^2*P(x_i=1)-(1*P(x_i=1))^2=p-p^2=p(1-p)$