Mixing
Solution for complex differentiation and integration.
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Following my question here, I am able to solve up to a point beyond which I would really appreciate any help (also if my current understanding is right).
Note that I have to find $f_T(t)$. The question starts with:
$$P(T<t) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
with $T, V$ are random variables with values $t,v$ denoting time and speed. $mathcalV$ is the range of $V$. $lambda$ is a rate (constant).
The final answer should be:
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$
where $$g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$$ and $$Q(x) = frac1sqrt2piint_x^infty e^fracu^22du$$
My Solution
Let velocity be a continous uniform distribution with range $[v_min,v_max] = [v_0,v_1]$, so above eq. would become:
$$P(T<t) = 1-int_v_0^v_1exp(-lambdapi v^2t^2)Big(frac1v_1-v_0Big)dv.$$
$$f_T(t) = -Big(frac1v_1-v_0Big)fracddtint_v_0^v_1exp(-lambdapi v^2t^2) dv.$$
using the property $int e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[texterf(sqrtpilambda tv_1) - texterf(sqrtpilambda tv_0)].$$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[Q(sqrtpilambda tv_0) - Q(sqrtpilambda tv_1)].$$
I cannot reach beyond this point. Any help would be much appreciated.
integration derivatives
asked Jul 31 at 23:31
Kashan
341110
add a comment |Â
up vote
0
down vote
favorite
Following my question here, I am able to solve up to a point beyond which I would really appreciate any help (also if my current understanding is right).
Note that I have to find $f_T(t)$. The question starts with:
$$P(T<t) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
with $T, V$ are random variables with values $t,v$ denoting time and speed. $mathcalV$ is the range of $V$. $lambda$ is a rate (constant).
The final answer should be:
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$
where $$g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$$ and $$Q(x) = frac1sqrt2piint_x^infty e^fracu^22du$$
My Solution
Let velocity be a continous uniform distribution with range $[v_min,v_max] = [v_0,v_1]$, so above eq. would become:
$$P(T<t) = 1-int_v_0^v_1exp(-lambdapi v^2t^2)Big(frac1v_1-v_0Big)dv.$$
$$f_T(t) = -Big(frac1v_1-v_0Big)fracddtint_v_0^v_1exp(-lambdapi v^2t^2) dv.$$
using the property $int e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[texterf(sqrtpilambda tv_1) - texterf(sqrtpilambda tv_0)].$$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[Q(sqrtpilambda tv_0) - Q(sqrtpilambda tv_1)].$$
I cannot reach beyond this point. Any help would be much appreciated.
integration derivatives
asked Jul 31 at 23:31
Kashan
341110
I am not familiar with random variables. Are you actually stuck at the point of differentiation?
– Szeto
Aug 1 at 2:05
Yes. I need the final solution by solving integral and derivate. for random variable, please consider $t,v$ as some variables
– Kashan
Aug 1 at 3:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Following my question here, I am able to solve up to a point beyond which I would really appreciate any help (also if my current understanding is right).
Note that I have to find $f_T(t)$. The question starts with:
$$P(T<t) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
with $T, V$ are random variables with values $t,v$ denoting time and speed. $mathcalV$ is the range of $V$. $lambda$ is a rate (constant).
The final answer should be:
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$
where $$g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$$ and $$Q(x) = frac1sqrt2piint_x^infty e^fracu^22du$$
My Solution
Let velocity be a continous uniform distribution with range $[v_min,v_max] = [v_0,v_1]$, so above eq. would become:
$$P(T<t) = 1-int_v_0^v_1exp(-lambdapi v^2t^2)Big(frac1v_1-v_0Big)dv.$$
$$f_T(t) = -Big(frac1v_1-v_0Big)fracddtint_v_0^v_1exp(-lambdapi v^2t^2) dv.$$
using the property $int e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[texterf(sqrtpilambda tv_1) - texterf(sqrtpilambda tv_0)].$$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[Q(sqrtpilambda tv_0) - Q(sqrtpilambda tv_1)].$$
I cannot reach beyond this point. Any help would be much appreciated.
integration derivatives
asked Jul 31 at 23:31
Kashan
341110
Following my question here, I am able to solve up to a point beyond which I would really appreciate any help (also if my current understanding is right).
Note that I have to find $f_T(t)$. The question starts with:
$$P(T<t) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
with $T, V$ are random variables with values $t,v$ denoting time and speed. $mathcalV$ is the range of $V$. $lambda$ is a rate (constant).
The final answer should be:
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$
where $$g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$$ and $$Q(x) = frac1sqrt2piint_x^infty e^fracu^22du$$
My Solution
Let velocity be a continous uniform distribution with range $[v_min,v_max] = [v_0,v_1]$, so above eq. would become:
$$P(T<t) = 1-int_v_0^v_1exp(-lambdapi v^2t^2)Big(frac1v_1-v_0Big)dv.$$
$$f_T(t) = -Big(frac1v_1-v_0Big)fracddtint_v_0^v_1exp(-lambdapi v^2t^2) dv.$$
using the property $int e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[texterf(sqrtpilambda tv_1) - texterf(sqrtpilambda tv_0)].$$
$$f_T(t) = -Big(frac1(v_1-v_0)2sqrtlambdaBig)fracddtfrac1t[Q(sqrtpilambda tv_0) - Q(sqrtpilambda tv_1)].$$
I cannot reach beyond this point. Any help would be much appreciated.
integration derivatives
asked Jul 31 at 23:31
Kashan
341110
edited Aug 1 at 5:20
asked Jul 31 at 23:31
Kashan
341110
asked Jul 31 at 23:31
Kashan
341110
asked Jul 31 at 23:31
Kashan
341110
341110
I am not familiar with random variables. Are you actually stuck at the point of differentiation?
– Szeto
Aug 1 at 2:05
Yes. I need the final solution by solving integral and derivate. for random variable, please consider $t,v$ as some variables
– Kashan
Aug 1 at 3:04
add a comment |Â
I am not familiar with random variables. Are you actually stuck at the point of differentiation?
– Szeto
Aug 1 at 2:05
Yes. I need the final solution by solving integral and derivate. for random variable, please consider $t,v$ as some variables
– Kashan
Aug 1 at 3:04
I am not familiar with random variables. Are you actually stuck at the point of differentiation?
– Szeto
Aug 1 at 2:05
I am not familiar with random variables. Are you actually stuck at the point of differentiation?
– Szeto
Aug 1 at 2:05
Yes. I need the final solution by solving integral and derivate. for random variable, please consider $t,v$ as some variables
– Kashan
Aug 1 at 3:04
Yes. I need the final solution by solving integral and derivate. for random variable, please consider $t,v$ as some variables
– Kashan
Aug 1 at 3:04
add a comment |Â
1 Answer
1
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up vote
2
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accepted
Using the property $e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
This is wrong. Look up the definition of error function.
Basically, you just need to find $fracddtP(t)$.
Let $$I(t) = int_v_0^v_1exp(-lambdapi v^2t^2)dv$$
(So, $P(t)=1-frac1v_1-v_0I(t)$)
Using the substitution $u=vt$,
$$I(t)=frac1tint_v_0t^v_1texp(-lambdapi u^2)du$$
By Fundamental Theorem of Calculus, this also equals
$$I(t)=frac1tleft(intexp(-lambdapi (v_1t)^2),,v_1dt-intexp(-lambdapi (v_0t)^2),,v_0dtright)$$
Applying chain rule,
$$I'(t)=fracI(t)+tv_1exp(-lambdapi (v_1t)^2)-tv_0exp(-lambdapi (v_0t)^2)t^2$$
The problem is basically solved, now it is a problem to fit in your notations.
Note that $$I(t)=sqrt2piQ(sqrt2lambdapiv_0t)-sqrt2piQ(sqrt2lambdapiv_1t)$$
...your notations are very messy...try to fit in yourself.
answered Aug 1 at 3:35
Szeto
3,8431421
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Using the property $e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
This is wrong. Look up the definition of error function.
Basically, you just need to find $fracddtP(t)$.
Let $$I(t) = int_v_0^v_1exp(-lambdapi v^2t^2)dv$$
(So, $P(t)=1-frac1v_1-v_0I(t)$)
Using the substitution $u=vt$,
$$I(t)=frac1tint_v_0t^v_1texp(-lambdapi u^2)du$$
By Fundamental Theorem of Calculus, this also equals
$$I(t)=frac1tleft(intexp(-lambdapi (v_1t)^2),,v_1dt-intexp(-lambdapi (v_0t)^2),,v_0dtright)$$
Applying chain rule,
$$I'(t)=fracI(t)+tv_1exp(-lambdapi (v_1t)^2)-tv_0exp(-lambdapi (v_0t)^2)t^2$$
The problem is basically solved, now it is a problem to fit in your notations.
Note that $$I(t)=sqrt2piQ(sqrt2lambdapiv_0t)-sqrt2piQ(sqrt2lambdapiv_1t)$$
...your notations are very messy...try to fit in yourself.
answered Aug 1 at 3:35
Szeto
3,8431421
add a comment |Â
up vote
2
down vote
accepted
Using the property $e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
This is wrong. Look up the definition of error function.
Basically, you just need to find $fracddtP(t)$.
Let $$I(t) = int_v_0^v_1exp(-lambdapi v^2t^2)dv$$
(So, $P(t)=1-frac1v_1-v_0I(t)$)
Using the substitution $u=vt$,
$$I(t)=frac1tint_v_0t^v_1texp(-lambdapi u^2)du$$
By Fundamental Theorem of Calculus, this also equals
$$I(t)=frac1tleft(intexp(-lambdapi (v_1t)^2),,v_1dt-intexp(-lambdapi (v_0t)^2),,v_0dtright)$$
Applying chain rule,
$$I'(t)=fracI(t)+tv_1exp(-lambdapi (v_1t)^2)-tv_0exp(-lambdapi (v_0t)^2)t^2$$
The problem is basically solved, now it is a problem to fit in your notations.
Note that $$I(t)=sqrt2piQ(sqrt2lambdapiv_0t)-sqrt2piQ(sqrt2lambdapiv_1t)$$
...your notations are very messy...try to fit in yourself.
answered Aug 1 at 3:35
Szeto
3,8431421
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Using the property $e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
This is wrong. Look up the definition of error function.
Basically, you just need to find $fracddtP(t)$.
Let $$I(t) = int_v_0^v_1exp(-lambdapi v^2t^2)dv$$
(So, $P(t)=1-frac1v_1-v_0I(t)$)
Using the substitution $u=vt$,
$$I(t)=frac1tint_v_0t^v_1texp(-lambdapi u^2)du$$
By Fundamental Theorem of Calculus, this also equals
$$I(t)=frac1tleft(intexp(-lambdapi (v_1t)^2),,v_1dt-intexp(-lambdapi (v_0t)^2),,v_0dtright)$$
Applying chain rule,
$$I'(t)=fracI(t)+tv_1exp(-lambdapi (v_1t)^2)-tv_0exp(-lambdapi (v_0t)^2)t^2$$
The problem is basically solved, now it is a problem to fit in your notations.
Note that $$I(t)=sqrt2piQ(sqrt2lambdapiv_0t)-sqrt2piQ(sqrt2lambdapiv_1t)$$
...your notations are very messy...try to fit in yourself.
answered Aug 1 at 3:35
Szeto
3,8431421
Using the property $e^-cx^2 = sqrtfracpi4c erf(sqrtcx)$
This is wrong. Look up the definition of error function.
Basically, you just need to find $fracddtP(t)$.
Let $$I(t) = int_v_0^v_1exp(-lambdapi v^2t^2)dv$$
(So, $P(t)=1-frac1v_1-v_0I(t)$)
Using the substitution $u=vt$,
$$I(t)=frac1tint_v_0t^v_1texp(-lambdapi u^2)du$$
By Fundamental Theorem of Calculus, this also equals
$$I(t)=frac1tleft(intexp(-lambdapi (v_1t)^2),,v_1dt-intexp(-lambdapi (v_0t)^2),,v_0dtright)$$
Applying chain rule,
$$I'(t)=fracI(t)+tv_1exp(-lambdapi (v_1t)^2)-tv_0exp(-lambdapi (v_0t)^2)t^2$$
The problem is basically solved, now it is a problem to fit in your notations.
Note that $$I(t)=sqrt2piQ(sqrt2lambdapiv_0t)-sqrt2piQ(sqrt2lambdapiv_1t)$$
...your notations are very messy...try to fit in yourself.
answered Aug 1 at 3:35
Szeto
3,8431421
answered Aug 1 at 3:35
Szeto
3,8431421
answered Aug 1 at 3:35
Szeto
3,8431421
answered Aug 1 at 3:35
Szeto
3,8431421
3,8431421
add a comment |Â
add a comment |Â
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I am not familiar with random variables. Are you actually stuck at the point of differentiation?
– Szeto
Aug 1 at 2:05
Yes. I need the final solution by solving integral and derivate. for random variable, please consider $t,v$ as some variables
– Kashan
Aug 1 at 3:04