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How does following transformations of random variables occur?
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While reading a paper related to the random mobility model, I am stuck with the following two questions. Let us start with the following equation.
$$P(L<l) = 1 - exp(-lambdapi l^2),$$ where L is the random variable and $l$ is the exact value. L denotes transition length and $L = VT$ $(V,T)$ are velocity and transition-time random varible with $v,t$ as respective values.
I understand that to find pdf of transition-time $T$ find f_T(t), we have to find CDF such that:
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-exp(-lambdapi v^2t^2).$$
But since $V$ is also a random variable, how come the authors say
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
where $mathcalV$ is the range of velocity and $P_V(.)$ is the velocity distribution. (where does $dP_V(v)$ comes from)
The second question is: If the velocity is uniformly distributed on $[v_min,v_max]$, I cannot understand how the pdf of transition time becomes
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$ with $g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$ and $Q(x) = frac1sqrt2piint_x^infty e^fracu^22$
probability-distributions random-variables transformation
asked Jul 31 at 12:36
Kashan
341110
add a comment |Â
up vote
1
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While reading a paper related to the random mobility model, I am stuck with the following two questions. Let us start with the following equation.
$$P(L<l) = 1 - exp(-lambdapi l^2),$$ where L is the random variable and $l$ is the exact value. L denotes transition length and $L = VT$ $(V,T)$ are velocity and transition-time random varible with $v,t$ as respective values.
I understand that to find pdf of transition-time $T$ find f_T(t), we have to find CDF such that:
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-exp(-lambdapi v^2t^2).$$
But since $V$ is also a random variable, how come the authors say
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
where $mathcalV$ is the range of velocity and $P_V(.)$ is the velocity distribution. (where does $dP_V(v)$ comes from)
The second question is: If the velocity is uniformly distributed on $[v_min,v_max]$, I cannot understand how the pdf of transition time becomes
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$ with $g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$ and $Q(x) = frac1sqrt2piint_x^infty e^fracu^22$
probability-distributions random-variables transformation
asked Jul 31 at 12:36
Kashan
341110
1
Assuming $L$ and $V$ are independent, indeed $$P(L<Vtmid V=v)=P(L<vt)=:h(v)$$ hence $$P(L<Vt)=E(P(L<Vtmid V))=E(h(V))=int_mathbb R h(v)dP_V(v)$$ where $P_V$ denotes the distribution of $V$. This is the formula in your post.
– Did
Jul 31 at 13:28
Thanks. That's a good hint.
– Kashan
Jul 31 at 22:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
While reading a paper related to the random mobility model, I am stuck with the following two questions. Let us start with the following equation.
$$P(L<l) = 1 - exp(-lambdapi l^2),$$ where L is the random variable and $l$ is the exact value. L denotes transition length and $L = VT$ $(V,T)$ are velocity and transition-time random varible with $v,t$ as respective values.
I understand that to find pdf of transition-time $T$ find f_T(t), we have to find CDF such that:
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-exp(-lambdapi v^2t^2).$$
But since $V$ is also a random variable, how come the authors say
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
where $mathcalV$ is the range of velocity and $P_V(.)$ is the velocity distribution. (where does $dP_V(v)$ comes from)
The second question is: If the velocity is uniformly distributed on $[v_min,v_max]$, I cannot understand how the pdf of transition time becomes
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$ with $g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$ and $Q(x) = frac1sqrt2piint_x^infty e^fracu^22$
probability-distributions random-variables transformation
asked Jul 31 at 12:36
Kashan
341110
While reading a paper related to the random mobility model, I am stuck with the following two questions. Let us start with the following equation.
$$P(L<l) = 1 - exp(-lambdapi l^2),$$ where L is the random variable and $l$ is the exact value. L denotes transition length and $L = VT$ $(V,T)$ are velocity and transition-time random varible with $v,t$ as respective values.
I understand that to find pdf of transition-time $T$ find f_T(t), we have to find CDF such that:
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-exp(-lambdapi v^2t^2).$$
But since $V$ is also a random variable, how come the authors say
$$P(T<t) = P(fracLV<t) = P(L<vt) = 1-int_mathcalVexp(-lambdapi v^2t^2)dP_V(v).$$
where $mathcalV$ is the range of velocity and $P_V(.)$ is the velocity distribution. (where does $dP_V(v)$ comes from)
The second question is: If the velocity is uniformly distributed on $[v_min,v_max]$, I cannot understand how the pdf of transition time becomes
$$f_T(t) = fracg(v_min) - g(v_max)(v_max - v_min)t, tgeq0$$ with $g(x) = xe^-lambda pi t^2x^2 + frac1sqrtlambdatQ(sqrt2pilambda t x)$ and $Q(x) = frac1sqrt2piint_x^infty e^fracu^22$
probability-distributions random-variables transformation
asked Jul 31 at 12:36
Kashan
341110
asked Jul 31 at 12:36
Kashan
341110
asked Jul 31 at 12:36
Kashan
341110
asked Jul 31 at 12:36
Kashan
341110
341110
1
Assuming $L$ and $V$ are independent, indeed $$P(L<Vtmid V=v)=P(L<vt)=:h(v)$$ hence $$P(L<Vt)=E(P(L<Vtmid V))=E(h(V))=int_mathbb R h(v)dP_V(v)$$ where $P_V$ denotes the distribution of $V$. This is the formula in your post.
– Did
Jul 31 at 13:28
Thanks. That's a good hint.
– Kashan
Jul 31 at 22:13
add a comment |Â
1
Assuming $L$ and $V$ are independent, indeed $$P(L<Vtmid V=v)=P(L<vt)=:h(v)$$ hence $$P(L<Vt)=E(P(L<Vtmid V))=E(h(V))=int_mathbb R h(v)dP_V(v)$$ where $P_V$ denotes the distribution of $V$. This is the formula in your post.
– Did
Jul 31 at 13:28
Thanks. That's a good hint.
– Kashan
Jul 31 at 22:13
1
1
Assuming $L$ and $V$ are independent, indeed $$P(L<Vtmid V=v)=P(L<vt)=:h(v)$$ hence $$P(L<Vt)=E(P(L<Vtmid V))=E(h(V))=int_mathbb R h(v)dP_V(v)$$ where $P_V$ denotes the distribution of $V$. This is the formula in your post.
– Did
Jul 31 at 13:28
Assuming $L$ and $V$ are independent, indeed $$P(L<Vtmid V=v)=P(L<vt)=:h(v)$$ hence $$P(L<Vt)=E(P(L<Vtmid V))=E(h(V))=int_mathbb R h(v)dP_V(v)$$ where $P_V$ denotes the distribution of $V$. This is the formula in your post.
– Did
Jul 31 at 13:28
Thanks. That's a good hint.
– Kashan
Jul 31 at 22:13
Thanks. That's a good hint.
– Kashan
Jul 31 at 22:13
add a comment |Â
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Assuming $L$ and $V$ are independent, indeed $$P(L<Vtmid V=v)=P(L<vt)=:h(v)$$ hence $$P(L<Vt)=E(P(L<Vtmid V))=E(h(V))=int_mathbb R h(v)dP_V(v)$$ where $P_V$ denotes the distribution of $V$. This is the formula in your post.
– Did
Jul 31 at 13:28
Thanks. That's a good hint.
– Kashan
Jul 31 at 22:13