Mixing
finding the variance of a random variable t, which depends on another random variable x.
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
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
add a comment |
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
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
– Alex
Dec 29 '18 at 20:59
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
– angryavian
Dec 29 '18 at 21:23
thank you, I fixed it. yes, I meant uniform distribution indeed
– Yanir Elm
Dec 29 '18 at 21:44
@YanirElm The question is wrong. $Et=infty$.
– Kavi Rama Murthy
Dec 29 '18 at 23:46
add a comment |
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
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
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
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
edited Dec 31 '18 at 20:12
Yanir Elm
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
asked Dec 29 '18 at 20:53
Yanir ElmYanir Elm
693
693
Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
– Alex
Dec 29 '18 at 20:59
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
– angryavian
Dec 29 '18 at 21:23
thank you, I fixed it. yes, I meant uniform distribution indeed
– Yanir Elm
Dec 29 '18 at 21:44
@YanirElm The question is wrong. $Et=infty$.
– Kavi Rama Murthy
Dec 29 '18 at 23:46
add a comment |
Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
– Alex
Dec 29 '18 at 20:59
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
– angryavian
Dec 29 '18 at 21:23
thank you, I fixed it. yes, I meant uniform distribution indeed
– Yanir Elm
Dec 29 '18 at 21:44
@YanirElm The question is wrong. $Et=infty$.
– Kavi Rama Murthy
Dec 29 '18 at 23:46
Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
– Alex
Dec 29 '18 at 20:59
Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
– Alex
Dec 29 '18 at 20:59
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
– angryavian
Dec 29 '18 at 21:23
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
– angryavian
Dec 29 '18 at 21:23
thank you, I fixed it. yes, I meant uniform distribution indeed
– Yanir Elm
Dec 29 '18 at 21:44
thank you, I fixed it. yes, I meant uniform distribution indeed
– Yanir Elm
Dec 29 '18 at 21:44
@YanirElm The question is wrong. $Et=infty$.
– Kavi Rama Murthy
Dec 29 '18 at 23:46
@YanirElm The question is wrong. $Et=infty$.
– Kavi Rama Murthy
Dec 29 '18 at 23:46
add a comment |
1 Answer
1
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oldest
votes
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 $mathsf E(T^2mid X)=dfrac{X^2}{3}$
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
– angryavian
Dec 29 '18 at 21:02
yes I mean E[1/x]
– Yanir Elm
Dec 29 '18 at 21:09
Why? Oh... I see... No.
– Graham Kemp
Dec 29 '18 at 21:11
maybe there's an error in the question itself..
– Yanir Elm
Dec 29 '18 at 21:48
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
– Graham Kemp
Dec 30 '18 at 6:51
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $mathsf E(T^2mid X)=dfrac{X^2}{3}$
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
– angryavian
Dec 29 '18 at 21:02
yes I mean E[1/x]
– Yanir Elm
Dec 29 '18 at 21:09
Why? Oh... I see... No.
– Graham Kemp
Dec 29 '18 at 21:11
maybe there's an error in the question itself..
– Yanir Elm
Dec 29 '18 at 21:48
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
– Graham Kemp
Dec 30 '18 at 6:51
|
show 1 more comment
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 $mathsf E(T^2mid X)=dfrac{X^2}{3}$
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
– angryavian
Dec 29 '18 at 21:02
yes I mean E[1/x]
– Yanir Elm
Dec 29 '18 at 21:09
Why? Oh... I see... No.
– Graham Kemp
Dec 29 '18 at 21:11
maybe there's an error in the question itself..
– Yanir Elm
Dec 29 '18 at 21:48
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
– Graham Kemp
Dec 30 '18 at 6:51
|
show 1 more comment
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 $mathsf E(T^2mid X)=dfrac{X^2}{3}$
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
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 $mathsf E(T^2mid X)=dfrac{X^2}{3}$
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
edited Dec 29 '18 at 21:10
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
answered Dec 29 '18 at 21:01
Graham KempGraham Kemp
84.8k43378
84.8k43378
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
– angryavian
Dec 29 '18 at 21:02
yes I mean E[1/x]
– Yanir Elm
Dec 29 '18 at 21:09
Why? Oh... I see... No.
– Graham Kemp
Dec 29 '18 at 21:11
maybe there's an error in the question itself..
– Yanir Elm
Dec 29 '18 at 21:48
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
– Graham Kemp
Dec 30 '18 at 6:51
|
show 1 more comment
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
– angryavian
Dec 29 '18 at 21:02
yes I mean E[1/x]
– Yanir Elm
Dec 29 '18 at 21:09
Why? Oh... I see... No.
– Graham Kemp
Dec 29 '18 at 21:11
maybe there's an error in the question itself..
– Yanir Elm
Dec 29 '18 at 21:48
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
– Graham Kemp
Dec 30 '18 at 6:51
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
– angryavian
Dec 29 '18 at 21:02
Isn't the OP trying to compute $E[1/X]$ and $E[1/X^2]$?
– angryavian
Dec 29 '18 at 21:02
yes I mean E[1/x]
– Yanir Elm
Dec 29 '18 at 21:09
yes I mean E[1/x]
– Yanir Elm
Dec 29 '18 at 21:09
Why? Oh... I see... No.
– Graham Kemp
Dec 29 '18 at 21:11
Why? Oh... I see... No.
– Graham Kemp
Dec 29 '18 at 21:11
maybe there's an error in the question itself..
– Yanir Elm
Dec 29 '18 at 21:48
maybe there's an error in the question itself..
– Yanir Elm
Dec 29 '18 at 21:48
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
– Graham Kemp
Dec 30 '18 at 6:51
No, the question is fine. The mistake is that $mathsf E(Tmid X)=X/2$ and not $1/(2X)$.
– Graham Kemp
Dec 30 '18 at 6:51
|
show 1 more comment
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Please see math.meta.stackexchange.com/questions/5020/… for help on formatting math so that it is easier to read on this site
– Alex
Dec 29 '18 at 20:59
What exactly does $U(1,1/x)$ mean? You seem to have mistakenly thought it was uniform on $[0, 1/x]$.
– angryavian
Dec 29 '18 at 21:23
thank you, I fixed it. yes, I meant uniform distribution indeed
– Yanir Elm
Dec 29 '18 at 21:44
@YanirElm The question is wrong. $Et=infty$.
– Kavi Rama Murthy
Dec 29 '18 at 23:46