Finding the Expectation and Variance, given the distribution function and density function for a continuous...












0














For this density function of a continuous random variable, X:



$$f(x) = begin{cases}
c & mbox{for } -1 ≤ x ≤ 1\
tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$



and calculating its distribution function:



$$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$



I am now unsure how to calculate the Expectation, Variance and Median.



Every time I try to using $int_{}^{} xf(x) dx$ for each of the intervals, I seem to get $EX = 0$



Any help will be greatly appreciated.
Thanks










share|cite|improve this question





























    0














    For this density function of a continuous random variable, X:



    $$f(x) = begin{cases}
    c & mbox{for } -1 ≤ x ≤ 1\
    tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$



    and calculating its distribution function:



    $$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$



    I am now unsure how to calculate the Expectation, Variance and Median.



    Every time I try to using $int_{}^{} xf(x) dx$ for each of the intervals, I seem to get $EX = 0$



    Any help will be greatly appreciated.
    Thanks










    share|cite|improve this question



























      0












      0








      0







      For this density function of a continuous random variable, X:



      $$f(x) = begin{cases}
      c & mbox{for } -1 ≤ x ≤ 1\
      tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$



      and calculating its distribution function:



      $$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$



      I am now unsure how to calculate the Expectation, Variance and Median.



      Every time I try to using $int_{}^{} xf(x) dx$ for each of the intervals, I seem to get $EX = 0$



      Any help will be greatly appreciated.
      Thanks










      share|cite|improve this question















      For this density function of a continuous random variable, X:



      $$f(x) = begin{cases}
      c & mbox{for } -1 ≤ x ≤ 1\
      tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$



      and calculating its distribution function:



      $$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$



      I am now unsure how to calculate the Expectation, Variance and Median.



      Every time I try to using $int_{}^{} xf(x) dx$ for each of the intervals, I seem to get $EX = 0$



      Any help will be greatly appreciated.
      Thanks







      probability-distributions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 27 at 0:16

























      asked Nov 27 at 0:03









      Mike Campbell

      62




      62






















          1 Answer
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          0














          Lets be more precise about things here.



          Firstly, the integrals you have written down for the distribution function $F(x)$ (which I assume you mean the Cumulative Distribution Function) are simply numbers. For example, $$int_1^infty frac{3}{8x^4} dx = frac 18$$



          is not something that depends on $x$. Instead, recall that the function $F(x)$ is defined as



          $$F(x) = int_{-infty}^x f(s)ds tag{1}$$



          whereas the function $f(x)$ satisfies



          $$int_{-infty}^infty f(s)ds = 1 tag{2}$$



          I assume you have already used $(2)$ and found the constant $c = frac 38$



          Thus when $x<-1$, using $(1)$, we have



          $$F(x) = int_{-infty}^x frac{3}{8s^4}ds = -frac{1}{8x^3}$$



          I will leave you to do the other two cases, but you should find that



          $$F(x) = begin{cases} -frac{1}{8x^3} & x leq -1 \ frac 38 x + frac 12 & -1<x<1 \ 1-frac{1}{8x^3} & x>1 end{cases}$$



          Next, as you have noted, we calculate the expectation using the formula



          $$Bbb E (X) = int_{-infty}^infty xf(x)dx$$



          But of course, for our function $f(x)$ we will have to split the domains of integration:



          begin{align}
          Bbb E (X) & = int_{-infty}^infty xf(x)dx \
          & = int_{-infty}^{-1} x cdot frac{3}{8x^4} dx + int_{-1}^1 x cdot frac 38 dx + int_1^infty x cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^3} dx + int_{-1}^1 frac {3x}{8} dx + int_1^infty frac{3}{8x^3} dx \
          & = -frac{3}{16} + 0 + frac{3}{16} \
          & = 0
          end{align}



          Indeed we find that the expected value is $0$, which is not surprising due to the fact that $f(x)$ is an even function.



          Next, the formula for the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2$$



          We already know that $Bbb E(X)=0$, so all that is left is to calculate $Bbb E(X^2)$. This is done in a similar fashion as before:



          begin{align}
          Bbb E (X^2) & = int_{-infty}^infty x^2f(x)dx \
          & = int_{-infty}^{-1} x^2 cdot frac{3}{8x^4} dx + int_{-1}^1 x^2 cdot frac 38 dx + int_1^infty x^2 cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^2} dx + int_{-1}^1 frac {3x^2}{8} dx + int_1^infty frac{3}{8x^2} dx \
          & = frac{3}{8} + frac 14 + frac{3}{8} \
          & = 1
          end{align}



          It follows that the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2 = 1 - (0)^2 = 1$$



          Finally, the Median amounts to finding the value of $x$ such that $F(x)=frac 12$. This is easily found to be $x = 0$.






          share|cite|improve this answer























          • Thank you so much for this. For some reason, I thought I was going wrong when I calculated the EX to be 0. But at least this confirms I was doing it correctly. Thank you
            – Mike Campbell
            Nov 27 at 11:23











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          1 Answer
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          0














          Lets be more precise about things here.



          Firstly, the integrals you have written down for the distribution function $F(x)$ (which I assume you mean the Cumulative Distribution Function) are simply numbers. For example, $$int_1^infty frac{3}{8x^4} dx = frac 18$$



          is not something that depends on $x$. Instead, recall that the function $F(x)$ is defined as



          $$F(x) = int_{-infty}^x f(s)ds tag{1}$$



          whereas the function $f(x)$ satisfies



          $$int_{-infty}^infty f(s)ds = 1 tag{2}$$



          I assume you have already used $(2)$ and found the constant $c = frac 38$



          Thus when $x<-1$, using $(1)$, we have



          $$F(x) = int_{-infty}^x frac{3}{8s^4}ds = -frac{1}{8x^3}$$



          I will leave you to do the other two cases, but you should find that



          $$F(x) = begin{cases} -frac{1}{8x^3} & x leq -1 \ frac 38 x + frac 12 & -1<x<1 \ 1-frac{1}{8x^3} & x>1 end{cases}$$



          Next, as you have noted, we calculate the expectation using the formula



          $$Bbb E (X) = int_{-infty}^infty xf(x)dx$$



          But of course, for our function $f(x)$ we will have to split the domains of integration:



          begin{align}
          Bbb E (X) & = int_{-infty}^infty xf(x)dx \
          & = int_{-infty}^{-1} x cdot frac{3}{8x^4} dx + int_{-1}^1 x cdot frac 38 dx + int_1^infty x cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^3} dx + int_{-1}^1 frac {3x}{8} dx + int_1^infty frac{3}{8x^3} dx \
          & = -frac{3}{16} + 0 + frac{3}{16} \
          & = 0
          end{align}



          Indeed we find that the expected value is $0$, which is not surprising due to the fact that $f(x)$ is an even function.



          Next, the formula for the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2$$



          We already know that $Bbb E(X)=0$, so all that is left is to calculate $Bbb E(X^2)$. This is done in a similar fashion as before:



          begin{align}
          Bbb E (X^2) & = int_{-infty}^infty x^2f(x)dx \
          & = int_{-infty}^{-1} x^2 cdot frac{3}{8x^4} dx + int_{-1}^1 x^2 cdot frac 38 dx + int_1^infty x^2 cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^2} dx + int_{-1}^1 frac {3x^2}{8} dx + int_1^infty frac{3}{8x^2} dx \
          & = frac{3}{8} + frac 14 + frac{3}{8} \
          & = 1
          end{align}



          It follows that the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2 = 1 - (0)^2 = 1$$



          Finally, the Median amounts to finding the value of $x$ such that $F(x)=frac 12$. This is easily found to be $x = 0$.






          share|cite|improve this answer























          • Thank you so much for this. For some reason, I thought I was going wrong when I calculated the EX to be 0. But at least this confirms I was doing it correctly. Thank you
            – Mike Campbell
            Nov 27 at 11:23
















          0














          Lets be more precise about things here.



          Firstly, the integrals you have written down for the distribution function $F(x)$ (which I assume you mean the Cumulative Distribution Function) are simply numbers. For example, $$int_1^infty frac{3}{8x^4} dx = frac 18$$



          is not something that depends on $x$. Instead, recall that the function $F(x)$ is defined as



          $$F(x) = int_{-infty}^x f(s)ds tag{1}$$



          whereas the function $f(x)$ satisfies



          $$int_{-infty}^infty f(s)ds = 1 tag{2}$$



          I assume you have already used $(2)$ and found the constant $c = frac 38$



          Thus when $x<-1$, using $(1)$, we have



          $$F(x) = int_{-infty}^x frac{3}{8s^4}ds = -frac{1}{8x^3}$$



          I will leave you to do the other two cases, but you should find that



          $$F(x) = begin{cases} -frac{1}{8x^3} & x leq -1 \ frac 38 x + frac 12 & -1<x<1 \ 1-frac{1}{8x^3} & x>1 end{cases}$$



          Next, as you have noted, we calculate the expectation using the formula



          $$Bbb E (X) = int_{-infty}^infty xf(x)dx$$



          But of course, for our function $f(x)$ we will have to split the domains of integration:



          begin{align}
          Bbb E (X) & = int_{-infty}^infty xf(x)dx \
          & = int_{-infty}^{-1} x cdot frac{3}{8x^4} dx + int_{-1}^1 x cdot frac 38 dx + int_1^infty x cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^3} dx + int_{-1}^1 frac {3x}{8} dx + int_1^infty frac{3}{8x^3} dx \
          & = -frac{3}{16} + 0 + frac{3}{16} \
          & = 0
          end{align}



          Indeed we find that the expected value is $0$, which is not surprising due to the fact that $f(x)$ is an even function.



          Next, the formula for the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2$$



          We already know that $Bbb E(X)=0$, so all that is left is to calculate $Bbb E(X^2)$. This is done in a similar fashion as before:



          begin{align}
          Bbb E (X^2) & = int_{-infty}^infty x^2f(x)dx \
          & = int_{-infty}^{-1} x^2 cdot frac{3}{8x^4} dx + int_{-1}^1 x^2 cdot frac 38 dx + int_1^infty x^2 cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^2} dx + int_{-1}^1 frac {3x^2}{8} dx + int_1^infty frac{3}{8x^2} dx \
          & = frac{3}{8} + frac 14 + frac{3}{8} \
          & = 1
          end{align}



          It follows that the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2 = 1 - (0)^2 = 1$$



          Finally, the Median amounts to finding the value of $x$ such that $F(x)=frac 12$. This is easily found to be $x = 0$.






          share|cite|improve this answer























          • Thank you so much for this. For some reason, I thought I was going wrong when I calculated the EX to be 0. But at least this confirms I was doing it correctly. Thank you
            – Mike Campbell
            Nov 27 at 11:23














          0












          0








          0






          Lets be more precise about things here.



          Firstly, the integrals you have written down for the distribution function $F(x)$ (which I assume you mean the Cumulative Distribution Function) are simply numbers. For example, $$int_1^infty frac{3}{8x^4} dx = frac 18$$



          is not something that depends on $x$. Instead, recall that the function $F(x)$ is defined as



          $$F(x) = int_{-infty}^x f(s)ds tag{1}$$



          whereas the function $f(x)$ satisfies



          $$int_{-infty}^infty f(s)ds = 1 tag{2}$$



          I assume you have already used $(2)$ and found the constant $c = frac 38$



          Thus when $x<-1$, using $(1)$, we have



          $$F(x) = int_{-infty}^x frac{3}{8s^4}ds = -frac{1}{8x^3}$$



          I will leave you to do the other two cases, but you should find that



          $$F(x) = begin{cases} -frac{1}{8x^3} & x leq -1 \ frac 38 x + frac 12 & -1<x<1 \ 1-frac{1}{8x^3} & x>1 end{cases}$$



          Next, as you have noted, we calculate the expectation using the formula



          $$Bbb E (X) = int_{-infty}^infty xf(x)dx$$



          But of course, for our function $f(x)$ we will have to split the domains of integration:



          begin{align}
          Bbb E (X) & = int_{-infty}^infty xf(x)dx \
          & = int_{-infty}^{-1} x cdot frac{3}{8x^4} dx + int_{-1}^1 x cdot frac 38 dx + int_1^infty x cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^3} dx + int_{-1}^1 frac {3x}{8} dx + int_1^infty frac{3}{8x^3} dx \
          & = -frac{3}{16} + 0 + frac{3}{16} \
          & = 0
          end{align}



          Indeed we find that the expected value is $0$, which is not surprising due to the fact that $f(x)$ is an even function.



          Next, the formula for the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2$$



          We already know that $Bbb E(X)=0$, so all that is left is to calculate $Bbb E(X^2)$. This is done in a similar fashion as before:



          begin{align}
          Bbb E (X^2) & = int_{-infty}^infty x^2f(x)dx \
          & = int_{-infty}^{-1} x^2 cdot frac{3}{8x^4} dx + int_{-1}^1 x^2 cdot frac 38 dx + int_1^infty x^2 cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^2} dx + int_{-1}^1 frac {3x^2}{8} dx + int_1^infty frac{3}{8x^2} dx \
          & = frac{3}{8} + frac 14 + frac{3}{8} \
          & = 1
          end{align}



          It follows that the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2 = 1 - (0)^2 = 1$$



          Finally, the Median amounts to finding the value of $x$ such that $F(x)=frac 12$. This is easily found to be $x = 0$.






          share|cite|improve this answer














          Lets be more precise about things here.



          Firstly, the integrals you have written down for the distribution function $F(x)$ (which I assume you mean the Cumulative Distribution Function) are simply numbers. For example, $$int_1^infty frac{3}{8x^4} dx = frac 18$$



          is not something that depends on $x$. Instead, recall that the function $F(x)$ is defined as



          $$F(x) = int_{-infty}^x f(s)ds tag{1}$$



          whereas the function $f(x)$ satisfies



          $$int_{-infty}^infty f(s)ds = 1 tag{2}$$



          I assume you have already used $(2)$ and found the constant $c = frac 38$



          Thus when $x<-1$, using $(1)$, we have



          $$F(x) = int_{-infty}^x frac{3}{8s^4}ds = -frac{1}{8x^3}$$



          I will leave you to do the other two cases, but you should find that



          $$F(x) = begin{cases} -frac{1}{8x^3} & x leq -1 \ frac 38 x + frac 12 & -1<x<1 \ 1-frac{1}{8x^3} & x>1 end{cases}$$



          Next, as you have noted, we calculate the expectation using the formula



          $$Bbb E (X) = int_{-infty}^infty xf(x)dx$$



          But of course, for our function $f(x)$ we will have to split the domains of integration:



          begin{align}
          Bbb E (X) & = int_{-infty}^infty xf(x)dx \
          & = int_{-infty}^{-1} x cdot frac{3}{8x^4} dx + int_{-1}^1 x cdot frac 38 dx + int_1^infty x cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^3} dx + int_{-1}^1 frac {3x}{8} dx + int_1^infty frac{3}{8x^3} dx \
          & = -frac{3}{16} + 0 + frac{3}{16} \
          & = 0
          end{align}



          Indeed we find that the expected value is $0$, which is not surprising due to the fact that $f(x)$ is an even function.



          Next, the formula for the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2$$



          We already know that $Bbb E(X)=0$, so all that is left is to calculate $Bbb E(X^2)$. This is done in a similar fashion as before:



          begin{align}
          Bbb E (X^2) & = int_{-infty}^infty x^2f(x)dx \
          & = int_{-infty}^{-1} x^2 cdot frac{3}{8x^4} dx + int_{-1}^1 x^2 cdot frac 38 dx + int_1^infty x^2 cdot frac{3}{8x^4} dx \
          & = int_{-infty}^{-1} frac{3}{8x^2} dx + int_{-1}^1 frac {3x^2}{8} dx + int_1^infty frac{3}{8x^2} dx \
          & = frac{3}{8} + frac 14 + frac{3}{8} \
          & = 1
          end{align}



          It follows that the variance is



          $$text{Var}(X) = Bbb E(X^2) - big[Bbb E(X)big]^2 = 1 - (0)^2 = 1$$



          Finally, the Median amounts to finding the value of $x$ such that $F(x)=frac 12$. This is easily found to be $x = 0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 at 0:43

























          answered Nov 27 at 0:35









          glowstonetrees

          2,285317




          2,285317












          • Thank you so much for this. For some reason, I thought I was going wrong when I calculated the EX to be 0. But at least this confirms I was doing it correctly. Thank you
            – Mike Campbell
            Nov 27 at 11:23


















          • Thank you so much for this. For some reason, I thought I was going wrong when I calculated the EX to be 0. But at least this confirms I was doing it correctly. Thank you
            – Mike Campbell
            Nov 27 at 11:23
















          Thank you so much for this. For some reason, I thought I was going wrong when I calculated the EX to be 0. But at least this confirms I was doing it correctly. Thank you
          – Mike Campbell
          Nov 27 at 11:23




          Thank you so much for this. For some reason, I thought I was going wrong when I calculated the EX to be 0. But at least this confirms I was doing it correctly. Thank you
          – Mike Campbell
          Nov 27 at 11:23


















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