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The probability density function of a random variable \mathrm{X}  is given by  \mathrm{f(x)=2 x}  for  \mathrm{0<x<1} . Find the mean and variance of \mathrm{x}.

Option: 1

\mathrm{ 2 / 3,1 / 18}


Option: 2

1 / 3,1 / 18


Option: 3

2 / 3,2 / 18


Option: 4

1 / 3,3 / 18


Answers (1)

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Solution : The mean \mathrm{ (\mu)} can be calculated as follows:

\mathrm{ \mu=\int\left[x^* f(x)\right] d x}

where the integral is taken over the range 0 to 1 , and  \mathrm{ f(x)=2 x}
\mathrm{ \mu=\int\left[x^{\star} 2 x\right] d x}  from  0  to  1 
\mathrm{ \mu=2 / 3 }
So, the mean (expected value) of the random variable \mathrm{X} is  \mathrm{ 2 / 3}
Similarly, the variance \mathrm{\left(\sigma^{\wedge} 2\right)}  can be calculated as follows:


\mathrm{\\mu=\int\left[x^* f(x)\right] d x}

where the integral is taken over the range 0 to 1 , and  \mathrm{f(x)=2 x}
\mathrm{ \mu=\int\left[x^{\star} 2 x\right] d x } from  0 to 1 
\mathrm{ \mu=2 / 3 }
So, the mean (expected value) of the random variable \mathrm{ \mathrm{X} }  is  \mathrm{ 2 / 3 }
Similarly, the variance\mathrm{ \left(\sigma^{\wedge} 2\right) } can be calculated as follows: 

\mathrm{ \sigma^{\wedge} 2=\int\left[(x-\mu)^{\wedge} 2 * f(x)\right] d x }
Now, integrate \mathrm{ (x-2 / 3)^{\wedge} 2 * 2 x }  with respect to \mathrm{x}:
\mathrm{ \sigma^{\wedge} 2=2 * \int\left[x^{\wedge} 3-(4 / 3) x^{\wedge} 2+(4 / 9) x\right] d x }

Now, integrate each term separately:

\mathrm{ \sigma^{\wedge} 2=2^{\star}\left[\left(x^{\wedge} 4 / 4\right)-(4 / 9) x^{\wedge} 3+(2 / 9) x^{\wedge} 2\right] \text { from } 0 \text { to } 1 }
\mathrm{ \sigma^{\wedge} 2=2^*[(1 / 4)-(4 / 9)+(2 / 9)]-2^{\star}[0] \\ }
\mathrm{\sigma^{\wedge} 2=2^{\star}[1 / 4-2 / 9] \\ }
\mathrm{ \sigma^{\wedge} 2=2^*[(9 / 36)-(8 / 36)] \\ }
\mathrm{\sigma^{\wedge} 2=2^*(1 / 36) }
\mathrm{ \sigma^{\wedge} 2=2 / 36 }
\mathrm{ \sigma^{\wedge} 2=1 / 18 }

So, the variance of the random variable  X  is  1 / 18 .

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manish

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