The mean value of sample moment with order k











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I've this problem of statistics that I can't resolve. So I hope that someone can halp me.



The problem is this:



Let sampling moments $overline{X_n^k}=frac{1}{n}sum_{i=1}^nX_i^k$, where for $k=1$ $overline{X_n^k}=overline{X_n}$.



Show that $E(overline{X_n^k})=E(X)$, where $E(Y)$ is the mean value of $Y$.



I did this:



$E(overline{X_n^k})=E(frac{1}{n}sum_{i=1}^nX_i^k)=frac{1}{n}E(sum_{i=1}^nX_i^k)$



I think that for ID of $X_i$, I can write:



$frac{1}{n}E(sum_{i=1}^nX_i^k)=frac{1}{n}*nE(X_i^k)=E(X_i^k)$.



Is this correct? And how can I go on? I thought to use the moment generating functions, but I don't know use it.



Thank you for your help.










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  • Well, $E(overline{X^k})=E(X)$ certainly doesn't hold in general. This, if at all, can only be true for particular value distributions, so we need more information on those.
    – Andreas
    21 hours ago










  • Did you intend to say "Show that $Eleft(overline{X_n^k}right)=Eleft(X^kright)$" with a $,^k$ on the right hand side?
    – Henry
    16 hours ago












  • I think that $X_i$ have random variable generative $Xsim N(mu,sigma ^2)$. My teacher didn't specify it, for this reason I didn't write it. But before of this exercise we treated this random variable.
    – user502400
    44 mins ago










  • I don't know Henry. This is the text that my teacher gave us, but she may have been wrong.
    – user502400
    39 mins ago

















up vote
0
down vote

favorite












I've this problem of statistics that I can't resolve. So I hope that someone can halp me.



The problem is this:



Let sampling moments $overline{X_n^k}=frac{1}{n}sum_{i=1}^nX_i^k$, where for $k=1$ $overline{X_n^k}=overline{X_n}$.



Show that $E(overline{X_n^k})=E(X)$, where $E(Y)$ is the mean value of $Y$.



I did this:



$E(overline{X_n^k})=E(frac{1}{n}sum_{i=1}^nX_i^k)=frac{1}{n}E(sum_{i=1}^nX_i^k)$



I think that for ID of $X_i$, I can write:



$frac{1}{n}E(sum_{i=1}^nX_i^k)=frac{1}{n}*nE(X_i^k)=E(X_i^k)$.



Is this correct? And how can I go on? I thought to use the moment generating functions, but I don't know use it.



Thank you for your help.










share|cite|improve this question






















  • Well, $E(overline{X^k})=E(X)$ certainly doesn't hold in general. This, if at all, can only be true for particular value distributions, so we need more information on those.
    – Andreas
    21 hours ago










  • Did you intend to say "Show that $Eleft(overline{X_n^k}right)=Eleft(X^kright)$" with a $,^k$ on the right hand side?
    – Henry
    16 hours ago












  • I think that $X_i$ have random variable generative $Xsim N(mu,sigma ^2)$. My teacher didn't specify it, for this reason I didn't write it. But before of this exercise we treated this random variable.
    – user502400
    44 mins ago










  • I don't know Henry. This is the text that my teacher gave us, but she may have been wrong.
    – user502400
    39 mins ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I've this problem of statistics that I can't resolve. So I hope that someone can halp me.



The problem is this:



Let sampling moments $overline{X_n^k}=frac{1}{n}sum_{i=1}^nX_i^k$, where for $k=1$ $overline{X_n^k}=overline{X_n}$.



Show that $E(overline{X_n^k})=E(X)$, where $E(Y)$ is the mean value of $Y$.



I did this:



$E(overline{X_n^k})=E(frac{1}{n}sum_{i=1}^nX_i^k)=frac{1}{n}E(sum_{i=1}^nX_i^k)$



I think that for ID of $X_i$, I can write:



$frac{1}{n}E(sum_{i=1}^nX_i^k)=frac{1}{n}*nE(X_i^k)=E(X_i^k)$.



Is this correct? And how can I go on? I thought to use the moment generating functions, but I don't know use it.



Thank you for your help.










share|cite|improve this question













I've this problem of statistics that I can't resolve. So I hope that someone can halp me.



The problem is this:



Let sampling moments $overline{X_n^k}=frac{1}{n}sum_{i=1}^nX_i^k$, where for $k=1$ $overline{X_n^k}=overline{X_n}$.



Show that $E(overline{X_n^k})=E(X)$, where $E(Y)$ is the mean value of $Y$.



I did this:



$E(overline{X_n^k})=E(frac{1}{n}sum_{i=1}^nX_i^k)=frac{1}{n}E(sum_{i=1}^nX_i^k)$



I think that for ID of $X_i$, I can write:



$frac{1}{n}E(sum_{i=1}^nX_i^k)=frac{1}{n}*nE(X_i^k)=E(X_i^k)$.



Is this correct? And how can I go on? I thought to use the moment generating functions, but I don't know use it.



Thank you for your help.







statistics means sampling






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asked 21 hours ago









user502400

214




214












  • Well, $E(overline{X^k})=E(X)$ certainly doesn't hold in general. This, if at all, can only be true for particular value distributions, so we need more information on those.
    – Andreas
    21 hours ago










  • Did you intend to say "Show that $Eleft(overline{X_n^k}right)=Eleft(X^kright)$" with a $,^k$ on the right hand side?
    – Henry
    16 hours ago












  • I think that $X_i$ have random variable generative $Xsim N(mu,sigma ^2)$. My teacher didn't specify it, for this reason I didn't write it. But before of this exercise we treated this random variable.
    – user502400
    44 mins ago










  • I don't know Henry. This is the text that my teacher gave us, but she may have been wrong.
    – user502400
    39 mins ago




















  • Well, $E(overline{X^k})=E(X)$ certainly doesn't hold in general. This, if at all, can only be true for particular value distributions, so we need more information on those.
    – Andreas
    21 hours ago










  • Did you intend to say "Show that $Eleft(overline{X_n^k}right)=Eleft(X^kright)$" with a $,^k$ on the right hand side?
    – Henry
    16 hours ago












  • I think that $X_i$ have random variable generative $Xsim N(mu,sigma ^2)$. My teacher didn't specify it, for this reason I didn't write it. But before of this exercise we treated this random variable.
    – user502400
    44 mins ago










  • I don't know Henry. This is the text that my teacher gave us, but she may have been wrong.
    – user502400
    39 mins ago


















Well, $E(overline{X^k})=E(X)$ certainly doesn't hold in general. This, if at all, can only be true for particular value distributions, so we need more information on those.
– Andreas
21 hours ago




Well, $E(overline{X^k})=E(X)$ certainly doesn't hold in general. This, if at all, can only be true for particular value distributions, so we need more information on those.
– Andreas
21 hours ago












Did you intend to say "Show that $Eleft(overline{X_n^k}right)=Eleft(X^kright)$" with a $,^k$ on the right hand side?
– Henry
16 hours ago






Did you intend to say "Show that $Eleft(overline{X_n^k}right)=Eleft(X^kright)$" with a $,^k$ on the right hand side?
– Henry
16 hours ago














I think that $X_i$ have random variable generative $Xsim N(mu,sigma ^2)$. My teacher didn't specify it, for this reason I didn't write it. But before of this exercise we treated this random variable.
– user502400
44 mins ago




I think that $X_i$ have random variable generative $Xsim N(mu,sigma ^2)$. My teacher didn't specify it, for this reason I didn't write it. But before of this exercise we treated this random variable.
– user502400
44 mins ago












I don't know Henry. This is the text that my teacher gave us, but she may have been wrong.
– user502400
39 mins ago






I don't know Henry. This is the text that my teacher gave us, but she may have been wrong.
– user502400
39 mins ago

















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