$f(x)=k x^{x} e^{-x}$ for x>0 and x=0 otherwise.

Find:

(i) k

(ii) Mean

(iii) Variance

(iv) MGF

(v) CDF of x

(vi) P[0 < x < 1]

Solution:

$f(x)=k x^{2}e^{-x}$

i) $\int_{0}^{\infty} kx^{2}e^{-x} \ dx=1$

$k\int_{0}^{\infty} x^{2}e^{-x} \ dx=1$

$k \left[ x^{2} . \cfrac{e^{-x}}{-1} - 2x\cfrac{e^{-x}}{(-1)(-1)}+\cfrac{2e^{-x}}{1(-1)} \right]_{0}^{\infty}=1$

$k[-x^{2}e^{-x}-2xe^{-x}-2e^{-x}]_{0}^{\infty}=1$

$k[[0-0-0]-[0-0-2(1)]]=1$

$k[2]=1$

$\therefore k=\cfrac{1}{2}$

ii) $\bar{X}=\mu_{1}=\int _{0}^{\infty} xf(x)dx=\int_{0}^{\infty} x.kx^{2}e^{-x} \ dx$

$\mu_{1}=\bar{X}=\int_{0}^{\infty} \cfrac{1}{2}x^{3}e^{-x} \ dx= \cfrac{1}{2} \int_{0}^{\infty} x^{3}e^{-x} \ dx = \cfrac{1}{2} \left[ \cfrac{x^{3}e^{-x}}{-1}-\cfrac{3x^{2}e^{-x}}{1} +\cfrac{6xe^{-x}}{-1} -\cfrac{6e^{-x}}{1} \right]_{0}^{\infty}$

$\mu_{1}=\bar{X}= \cfrac{1}{2} \left[ -x^{3}e^{-x}-3x^{2}e^{-x}-6xe^{-x} -6e^{-x} \right]_{0}^{\infty}= \cfrac{1}{2} [[0-0-0-0]-[0-0-0-6(1)]]$

$\therefore \mu_{1}=\bar{X}= \cfrac{1}{2}[6]=3$

iii) $\mu_{2}=\int _{0}^{\infty} x^{2}f(x)dx=\int_{0}^{\infty} x^{2}.kx^{2}e^{-x} \ dx$

$\mu_{2}=\int_{0}^{\infty} \cfrac{1}{2}x^{4}e^{-x} \ dx= \cfrac{1}{2} \int_{0}^{\infty} x^{4}e^{-x} \ dx = \cfrac{1}{2} \left[ \cfrac{x^{4}e^{-x}}{-1}-\cfrac{4x^{3}e^{-x}}{1} +\cfrac{12x^{2}e^{-x}}{-1} -\cfrac{24xe^{-x}}{1} +\cfrac{24e^{-x}}{-1} \right]_{0}^{\infty}$

$\mu_{2}= \cfrac{1}{2} \left[ -x^{4}e^{-x}-4x^{3}e^{-x}-12x^{2}e^{-x}-24xe^{-x} -24e^{-x} \right]_{0}^{\infty}= \cfrac{1}{2} [[-0-0-0-0-0]-[-0-0-0-0-24(1)]]$

$\therefore \mu_{2}= \cfrac{1}{2}[24]=12$

$\therefore $ Variance$=\mu_{2}-\mu_{1}^{2}=12-(3)^{2}=12-9=3$

iv) $M_{0}(t)=E(e^{tx})= \int_{0}^{\infty}e^{tx}.kx^{2}e^{-x} \ dx= \int_{0}^{\infty}e^{tx}.\cfrac{1}{2}x^{2}e^{-x} \ dx= \cfrac{1}{2} \int_{0}^{\infty}x^{2}e^{(tx-x)} \ dx$

$=\cfrac{1}{2} \left[ \cfrac{x^{2}e^{(tx-x)}}{(t-1)} -\cfrac{2xe^{(tx-x)}}{(t-1)^{2}} + \cfrac{2e^{(tx-x)}}{(t-1)^{3}} \right]_{0}^{\infty}= \cfrac{1}{2} \left[ \left[ 0-0+0\right] -\left[ 0-0+\cfrac{2}{(t-1)^{3}} \right]\right] = \cfrac{1}{2} \left[ \cfrac{-2}{(t-1)^{3}} \right]$

$\therefore M_{0}(t)= -\cfrac{1}{(t-1)^{3}}$

v) CDF

We have to verify, $\int_{-\infty}^{\infty}f(x) \ dx=1$

$\int_{-\infty}^{0} 0 \ dx+ \int_{0}^{\infty} kx^{2}e^{-x} \ dx=\int_{0}^{\infty} \cfrac{1}{2}x^{2}e^{-x} \ dx= \cfrac{1}{2} \int_{0}^{\infty} x^{2}e^{-x} \ dx =\cfrac{1}{2} \left[ x^{2} . \cfrac{e^{-x}}{-1} - 2x\cfrac{e^{-x}}{(-1)(-1)}+\cfrac{2e^{-x}}{1(-1)} \right]_{0}^{\infty}=\cfrac{1}{2}[-x^{2}e^{-x}-2xe^{-x}-2e^{-x}]_{0}^{\infty}=\cfrac{1}{2}[[0-0-0]-[-0-0-2(1)]]=\cfrac{1}{2}[2]=1$

$\therefore \int_{-\infty}^{\infty}f(x) \ dx=1$

Distribution function is already given by:

$F(x)=\int_{0}^{x}f(x) \ dx =\int_{0}^{x} kx^{2}e^{-x} \ dx=\cfrac{1}{2} \int_{0}^{x} x^{2}e^{-x} \ dx= \cfrac{1}{2}[-x^{2}e^{-x}-2xe^{-x}-2e^{-x}]_{0}^{x}=\cfrac{1}{2}[[-x^{2}e^{-x}-2xe^{-x}-2e^{-x}]-[0-0-2]]=\cfrac{1}{2}[-x^{2}e^{-x}-2xe^{-x}-2e^{-x}+2]$

vi) P[0 < x < 1]

$f(x)=kx^{2}e^{-x}$

$f(x)=\cfrac{1}{2}x^{2}e^{-x}$ --------(1)

P[0 < x < 1]$=F(1)-F(0)$

Put x=1 and x=0 in (1),

P[0 < x < 1]$=\cfrac{1}{2}(1)^{2}e^{-1}-\cfrac{1}{2}(0)^{2}e^{-x}=\cfrac{1}{2}e^{-1}= \cfrac{1}{2e}$