finding the variance of a random variable t, which depends on another random variable x.












0












$begingroup$


I know that (U stands for uniform distribution)$$ t|x sim U(0,frac{1}{x})$$
while x is a random variable itself, distributed exponentially:
$$xsim exp(lambda)$$



now i thought using the law of total expectation in order to find the variance of t:



$$E[t] = E[E[t|x]] = E[frac{1}{2x}]$$



and
$$ E[t^2]=E[E[t^2|x]] = E[frac{1}{12x^2} + frac{1}{(2x)^2}] = E[frac{1}{3x^2}] $$



and then calculating
$$ V[t]=E[t^2]-(E[t])^2=E[frac{1}{3x^2}]-E[frac{1}{2x}]$$



but knowing
$$ x sim exp(lambda)$$



each of these terms diverges..



am I wrong? is there any other way doing this?






probability statistics probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
    $endgroup$
    – Alex
    Dec 29 '18 at 20:59










  • $begingroup$
    What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:23










  • $begingroup$
    thank you, I fixed it. yes, I meant uniform distribution indeed
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:44












  • $begingroup$
    @YanirElm The question is wrong. $Et=infty$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:46
















0












$begingroup$


I know that (U stands for uniform distribution)$$ t|x sim U(0,frac{1}{x})$$
while x is a random variable itself, distributed exponentially:
$$xsim exp(lambda)$$



now i thought using the law of total expectation in order to find the variance of t:



$$E[t] = E[E[t|x]] = E[frac{1}{2x}]$$



and
$$ E[t^2]=E[E[t^2|x]] = E[frac{1}{12x^2} + frac{1}{(2x)^2}] = E[frac{1}{3x^2}] $$



and then calculating
$$ V[t]=E[t^2]-(E[t])^2=E[frac{1}{3x^2}]-E[frac{1}{2x}]$$



but knowing
$$ x sim exp(lambda)$$



each of these terms diverges..



am I wrong? is there any other way doing this?






probability statistics probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
    $endgroup$
    – Alex
    Dec 29 '18 at 20:59










  • $begingroup$
    What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:23










  • $begingroup$
    thank you, I fixed it. yes, I meant uniform distribution indeed
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:44












  • $begingroup$
    @YanirElm The question is wrong. $Et=infty$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:46














0












0








0





$begingroup$


I know that (U stands for uniform distribution)$$ t|x sim U(0,frac{1}{x})$$
while x is a random variable itself, distributed exponentially:
$$xsim exp(lambda)$$



now i thought using the law of total expectation in order to find the variance of t:



$$E[t] = E[E[t|x]] = E[frac{1}{2x}]$$



and
$$ E[t^2]=E[E[t^2|x]] = E[frac{1}{12x^2} + frac{1}{(2x)^2}] = E[frac{1}{3x^2}] $$



and then calculating
$$ V[t]=E[t^2]-(E[t])^2=E[frac{1}{3x^2}]-E[frac{1}{2x}]$$



but knowing
$$ x sim exp(lambda)$$



each of these terms diverges..



am I wrong? is there any other way doing this?






probability statistics probability-distributions







share|cite|improve this question











$endgroup$




I know that (U stands for uniform distribution)$$ t|x sim U(0,frac{1}{x})$$
while x is a random variable itself, distributed exponentially:
$$xsim exp(lambda)$$



now i thought using the law of total expectation in order to find the variance of t:



$$E[t] = E[E[t|x]] = E[frac{1}{2x}]$$



and
$$ E[t^2]=E[E[t^2|x]] = E[frac{1}{12x^2} + frac{1}{(2x)^2}] = E[frac{1}{3x^2}] $$



and then calculating
$$ V[t]=E[t^2]-(E[t])^2=E[frac{1}{3x^2}]-E[frac{1}{2x}]$$



but knowing
$$ x sim exp(lambda)$$



each of these terms diverges..



am I wrong? is there any other way doing this?






probability statistics probability-distributions




probability statistics probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 20:12







Yanir Elm

















asked Dec 29 '18 at 20:53









Yanir ElmYanir Elm

693




693












  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
    $endgroup$
    – Alex
    Dec 29 '18 at 20:59










  • $begingroup$
    What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:23










  • $begingroup$
    thank you, I fixed it. yes, I meant uniform distribution indeed
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:44












  • $begingroup$
    @YanirElm The question is wrong. $Et=infty$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:46


















  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
    $endgroup$
    – Alex
    Dec 29 '18 at 20:59










  • $begingroup$
    What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:23










  • $begingroup$
    thank you, I fixed it. yes, I meant uniform distribution indeed
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:44












  • $begingroup$
    @YanirElm The question is wrong. $Et=infty$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:46
















$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
$endgroup$
– Alex
Dec 29 '18 at 20:59




$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
$endgroup$
– Alex
Dec 29 '18 at 20:59












$begingroup$
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
$endgroup$
– angryavian
Dec 29 '18 at 21:23




$begingroup$
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
$endgroup$
– angryavian
Dec 29 '18 at 21:23












$begingroup$
thank you, I fixed it. yes, I meant uniform distribution indeed
$endgroup$
– Yanir Elm
Dec 29 '18 at 21:44






$begingroup$
thank you, I fixed it. yes, I meant uniform distribution indeed
$endgroup$
– Yanir Elm
Dec 29 '18 at 21:44














$begingroup$
@YanirElm The question is wrong. $Et=infty$.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:46




$begingroup$
@YanirElm The question is wrong. $Et=infty$.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:46










1 Answer
1






active

oldest

votes


















0












$begingroup$

The mean and variance of an exponential distribution do converge.   Because $Xsimmathcal{Exp}(lambda)$, therefore:$$begin{split}mathsf E(X)&= lambda^{-1}\mathsf E(X^2)-mathsf E(X)^2&=lambda^{-2}end{split}$$




Also since $Tmid Xsimmathcal U(1,tfrac 1X)$ therefore: $mathsf E(Tmid X)=dfrac X2$ and $mathsf E(T^2mid X)=dfrac{X^2}{3}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:02










  • $begingroup$
    yes I mean E[1/x]
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:09










  • $begingroup$
    Why? Oh... I see... No.
    $endgroup$
    – Graham Kemp
    Dec 29 '18 at 21:11










  • $begingroup$
    maybe there's an error in the question itself..
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:48










  • $begingroup$
    No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
    $endgroup$
    – Graham Kemp
    Dec 30 '18 at 6:51











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Post as a guest















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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The mean and variance of an exponential distribution do converge.   Because $Xsimmathcal{Exp}(lambda)$, therefore:$$begin{split}mathsf E(X)&= lambda^{-1}\mathsf E(X^2)-mathsf E(X)^2&=lambda^{-2}end{split}$$




Also since $Tmid Xsimmathcal U(1,tfrac 1X)$ therefore: $mathsf E(Tmid X)=dfrac X2$ and $mathsf E(T^2mid X)=dfrac{X^2}{3}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:02










  • $begingroup$
    yes I mean E[1/x]
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:09










  • $begingroup$
    Why? Oh... I see... No.
    $endgroup$
    – Graham Kemp
    Dec 29 '18 at 21:11










  • $begingroup$
    maybe there's an error in the question itself..
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:48










  • $begingroup$
    No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
    $endgroup$
    – Graham Kemp
    Dec 30 '18 at 6:51
















0












$begingroup$

The mean and variance of an exponential distribution do converge.   Because $Xsimmathcal{Exp}(lambda)$, therefore:$$begin{split}mathsf E(X)&= lambda^{-1}\mathsf E(X^2)-mathsf E(X)^2&=lambda^{-2}end{split}$$




Also since $Tmid Xsimmathcal U(1,tfrac 1X)$ therefore: $mathsf E(Tmid X)=dfrac X2$ and $mathsf E(T^2mid X)=dfrac{X^2}{3}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:02










  • $begingroup$
    yes I mean E[1/x]
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:09










  • $begingroup$
    Why? Oh... I see... No.
    $endgroup$
    – Graham Kemp
    Dec 29 '18 at 21:11










  • $begingroup$
    maybe there's an error in the question itself..
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:48










  • $begingroup$
    No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
    $endgroup$
    – Graham Kemp
    Dec 30 '18 at 6:51














0












0








0





$begingroup$

The mean and variance of an exponential distribution do converge.   Because $Xsimmathcal{Exp}(lambda)$, therefore:$$begin{split}mathsf E(X)&= lambda^{-1}\mathsf E(X^2)-mathsf E(X)^2&=lambda^{-2}end{split}$$




Also since $Tmid Xsimmathcal U(1,tfrac 1X)$ therefore: $mathsf E(Tmid X)=dfrac X2$ and $mathsf E(T^2mid X)=dfrac{X^2}{3}$






share|cite|improve this answer











$endgroup$



The mean and variance of an exponential distribution do converge.   Because $Xsimmathcal{Exp}(lambda)$, therefore:$$begin{split}mathsf E(X)&= lambda^{-1}\mathsf E(X^2)-mathsf E(X)^2&=lambda^{-2}end{split}$$




Also since $Tmid Xsimmathcal U(1,tfrac 1X)$ therefore: $mathsf E(Tmid X)=dfrac X2$ and $mathsf E(T^2mid X)=dfrac{X^2}{3}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 21:10

























answered Dec 29 '18 at 21:01









Graham KempGraham Kemp

87.1k43579




87.1k43579












  • $begingroup$
    Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:02










  • $begingroup$
    yes I mean E[1/x]
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:09










  • $begingroup$
    Why? Oh... I see... No.
    $endgroup$
    – Graham Kemp
    Dec 29 '18 at 21:11










  • $begingroup$
    maybe there's an error in the question itself..
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:48










  • $begingroup$
    No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
    $endgroup$
    – Graham Kemp
    Dec 30 '18 at 6:51


















  • $begingroup$
    Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:02










  • $begingroup$
    yes I mean E[1/x]
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:09










  • $begingroup$
    Why? Oh... I see... No.
    $endgroup$
    – Graham Kemp
    Dec 29 '18 at 21:11










  • $begingroup$
    maybe there's an error in the question itself..
    $endgroup$
    – Yanir Elm
    Dec 29 '18 at 21:48










  • $begingroup$
    No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
    $endgroup$
    – Graham Kemp
    Dec 30 '18 at 6:51
















$begingroup$
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
$endgroup$
– angryavian
Dec 29 '18 at 21:02




$begingroup$
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
$endgroup$
– angryavian
Dec 29 '18 at 21:02












$begingroup$
yes I mean E[1/x]
$endgroup$
– Yanir Elm
Dec 29 '18 at 21:09




$begingroup$
yes I mean E[1/x]
$endgroup$
– Yanir Elm
Dec 29 '18 at 21:09












$begingroup$
Why? Oh... I see... No.
$endgroup$
– Graham Kemp
Dec 29 '18 at 21:11




$begingroup$
Why? Oh... I see... No.
$endgroup$
– Graham Kemp
Dec 29 '18 at 21:11












$begingroup$
maybe there's an error in the question itself..
$endgroup$
– Yanir Elm
Dec 29 '18 at 21:48




$begingroup$
maybe there's an error in the question itself..
$endgroup$
– Yanir Elm
Dec 29 '18 at 21:48












$begingroup$
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
$endgroup$
– Graham Kemp
Dec 30 '18 at 6:51




$begingroup$
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
$endgroup$
– Graham Kemp
Dec 30 '18 at 6:51


















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