Mixing
Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.
$begingroup$
Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.
I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.
Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.
Could someone check if I'm good till now and maybe help me continue? Thanks!
integration probability-theory random-variables conditional-expectation expected-value
edited Jan 8 at 22:06
Batominovski
1
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
$endgroup$
add a comment |
$begingroup$
Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.
I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.
Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.
Could someone check if I'm good till now and maybe help me continue? Thanks!
integration probability-theory random-variables conditional-expectation expected-value
edited Jan 8 at 22:06
Batominovski
1
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
$endgroup$
1
$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05
$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28
1
$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46
$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39
add a comment |
$begingroup$
Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.
I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.
Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.
Could someone check if I'm good till now and maybe help me continue? Thanks!
integration probability-theory random-variables conditional-expectation expected-value
edited Jan 8 at 22:06
Batominovski
1
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
$endgroup$
Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.
I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.
Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.
Could someone check if I'm good till now and maybe help me continue? Thanks!
integration probability-theory random-variables conditional-expectation expected-value
integration probability-theory random-variables conditional-expectation expected-value
edited Jan 8 at 22:06
Batominovski
1
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
edited Jan 8 at 22:06
Batominovski
1
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
edited Jan 8 at 22:06
Batominovski
1
edited Jan 8 at 22:06
Batominovski
1
edited Jan 8 at 22:06
Batominovski
1
1
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
asked Jan 8 at 19:57
rapidracimrapidracim
1,6141319
1,6141319
1
$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05
$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28
1
$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46
$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39
add a comment |
1
$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05
$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28
1
$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46
$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39
1
1
$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05
$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05
$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28
$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28
1
1
$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46
$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46
$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39
$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
$$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
(Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
In other words,
$$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
This also shows that
$$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$
The required integral is
begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
&=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
&=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
Now
$$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
This gives
$$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$
$endgroup$
add a comment |
$begingroup$
You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
I calculated this result and got I=3.0591165.
In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$
Net result$=2.365969319$.
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
$endgroup$
$begingroup$
@ Batominovski What was the edit? I didn't see any changes!
$endgroup$
– herb steinberg
Jan 8 at 22:34
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
$$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
(Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
In other words,
$$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
This also shows that
$$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$
The required integral is
begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
&=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
&=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
Now
$$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
This gives
$$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$
$endgroup$
add a comment |
$begingroup$
You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
$$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
(Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
In other words,
$$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
This also shows that
$$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$
The required integral is
begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
&=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
&=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
Now
$$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
This gives
$$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$
$endgroup$
add a comment |
$begingroup$
You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
$$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
(Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
In other words,
$$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
This also shows that
$$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$
The required integral is
begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
&=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
&=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
Now
$$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
This gives
$$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$
$endgroup$
You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
$$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
(Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
In other words,
$$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
This also shows that
$$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$
The required integral is
begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
&=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
&=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
Now
$$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
This gives
$$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$
edited Jan 8 at 20:58
answered Jan 8 at 20:50
user614671
add a comment |
add a comment |
$begingroup$
You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
I calculated this result and got I=3.0591165.
In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$
Net result$=2.365969319$.
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
$endgroup$
$begingroup$
@ Batominovski What was the edit? I didn't see any changes!
$endgroup$
– herb steinberg
Jan 8 at 22:34
add a comment |
$begingroup$
You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
I calculated this result and got I=3.0591165.
In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$
Net result$=2.365969319$.
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
$endgroup$
$begingroup$
@ Batominovski What was the edit? I didn't see any changes!
$endgroup$
– herb steinberg
Jan 8 at 22:34
add a comment |
$begingroup$
You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
I calculated this result and got I=3.0591165.
In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$
Net result$=2.365969319$.
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
$endgroup$
You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
I calculated this result and got I=3.0591165.
In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$
Net result$=2.365969319$.
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
edited Jan 14 at 20:44
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
answered Jan 8 at 21:06
herb steinbergherb steinberg
2,5932310
2,5932310
$begingroup$
@ Batominovski What was the edit? I didn't see any changes!
$endgroup$
– herb steinberg
Jan 8 at 22:34
add a comment |
$begingroup$
@ Batominovski What was the edit? I didn't see any changes!
$endgroup$
– herb steinberg
Jan 8 at 22:34
$begingroup$
@ Batominovski What was the edit? I didn't see any changes!
$endgroup$
– herb steinberg
Jan 8 at 22:34
$begingroup$
@ Batominovski What was the edit? I didn't see any changes!
$endgroup$
– herb steinberg
Jan 8 at 22:34
add a comment |
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1
$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05
$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28
1
$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46
$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39