Mixing
Need help with transformation formula: how to get the pdf of $T$ given you know the pdf of $ln(T)$
$begingroup$
I know the answer is:
$$f_T(t) = frac{d}{dt} F_{ln t}(ln(t)) = f_{ln t}(ln t) frac{1}{t}$$
which you can get by using cdfs and taking the derivative.
$$F_T(t) = P(T le t) = P(ln(T) le ln(t)) = F_{ln t}(ln(t))$$
My question: I am having difficulty getting the answer using this alternative approach which uses the formula:
$$f_Y(y) = f_X(x) frac{dx}{dy} quad text{with} quad x = g^{-1}(y)$$
My attempt:
Here $Y = T$ and $X = ln T$ so we have
$$f_T(t) = f_{ln T}(x) frac{dx}{dt} $$
and now I have to find $x = g^{-1}(t)$ but at this point I'm stuck. I know $x = lnt$ but I'm having issues writing this out formally.
It's tripping me out because we go from $lnt$ to $t$ which is a simpler function. Thanks for your help and patience.
probability density-function
edited Jan 28 at 20:37
Bernard
123k741117
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
$endgroup$
|
show 1 more comment
$begingroup$
I know the answer is:
$$f_T(t) = frac{d}{dt} F_{ln t}(ln(t)) = f_{ln t}(ln t) frac{1}{t}$$
which you can get by using cdfs and taking the derivative.
$$F_T(t) = P(T le t) = P(ln(T) le ln(t)) = F_{ln t}(ln(t))$$
My question: I am having difficulty getting the answer using this alternative approach which uses the formula:
$$f_Y(y) = f_X(x) frac{dx}{dy} quad text{with} quad x = g^{-1}(y)$$
My attempt:
Here $Y = T$ and $X = ln T$ so we have
$$f_T(t) = f_{ln T}(x) frac{dx}{dt} $$
and now I have to find $x = g^{-1}(t)$ but at this point I'm stuck. I know $x = lnt$ but I'm having issues writing this out formally.
It's tripping me out because we go from $lnt$ to $t$ which is a simpler function. Thanks for your help and patience.
probability density-function
edited Jan 28 at 20:37
Bernard
123k741117
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
$endgroup$
$begingroup$
$dt=tdx = e^xdx$
$endgroup$
– phdmba7of12
Jan 28 at 19:17
$begingroup$
the pdf is defined such that $f(x)dx$ gives the probability that $x$ will be measured in between $x$ and $x+dx$
$endgroup$
– phdmba7of12
Jan 28 at 19:19
1
$begingroup$
@phdmba7of12 thanks for your help... hmm it seems $t=e^x$? How do you get to the conclusion that $x=g^{-1}(t) = lnt$? Thanks again!
$endgroup$
– HJ_beginner
Jan 28 at 19:24
$begingroup$
$e^{ln z}=z$ just exponentiate both sides of the $x = ln t$ equation
$endgroup$
– phdmba7of12
Jan 28 at 19:25
$begingroup$
one way might be to use delta function as kernel for the transform: $$F(x)=int_0^x f(x)dx$$ and $$F(ln t)=int_0^{ln t}f(ln t)dleft(ln t right)$$
$endgroup$
– phdmba7of12
Jan 28 at 19:27
|
show 1 more comment
$begingroup$
I know the answer is:
$$f_T(t) = frac{d}{dt} F_{ln t}(ln(t)) = f_{ln t}(ln t) frac{1}{t}$$
which you can get by using cdfs and taking the derivative.
$$F_T(t) = P(T le t) = P(ln(T) le ln(t)) = F_{ln t}(ln(t))$$
My question: I am having difficulty getting the answer using this alternative approach which uses the formula:
$$f_Y(y) = f_X(x) frac{dx}{dy} quad text{with} quad x = g^{-1}(y)$$
My attempt:
Here $Y = T$ and $X = ln T$ so we have
$$f_T(t) = f_{ln T}(x) frac{dx}{dt} $$
and now I have to find $x = g^{-1}(t)$ but at this point I'm stuck. I know $x = lnt$ but I'm having issues writing this out formally.
It's tripping me out because we go from $lnt$ to $t$ which is a simpler function. Thanks for your help and patience.
probability density-function
edited Jan 28 at 20:37
Bernard
123k741117
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
$endgroup$
I know the answer is:
$$f_T(t) = frac{d}{dt} F_{ln t}(ln(t)) = f_{ln t}(ln t) frac{1}{t}$$
which you can get by using cdfs and taking the derivative.
$$F_T(t) = P(T le t) = P(ln(T) le ln(t)) = F_{ln t}(ln(t))$$
My question: I am having difficulty getting the answer using this alternative approach which uses the formula:
$$f_Y(y) = f_X(x) frac{dx}{dy} quad text{with} quad x = g^{-1}(y)$$
My attempt:
Here $Y = T$ and $X = ln T$ so we have
$$f_T(t) = f_{ln T}(x) frac{dx}{dt} $$
and now I have to find $x = g^{-1}(t)$ but at this point I'm stuck. I know $x = lnt$ but I'm having issues writing this out formally.
It's tripping me out because we go from $lnt$ to $t$ which is a simpler function. Thanks for your help and patience.
probability density-function
probability density-function
edited Jan 28 at 20:37
Bernard
123k741117
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
edited Jan 28 at 20:37
Bernard
123k741117
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
edited Jan 28 at 20:37
Bernard
123k741117
edited Jan 28 at 20:37
Bernard
123k741117
edited Jan 28 at 20:37
Bernard
123k741117
123k741117
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
asked Jan 28 at 19:12
HJ_beginnerHJ_beginner
8951415
8951415
$begingroup$
$dt=tdx = e^xdx$
$endgroup$
– phdmba7of12
Jan 28 at 19:17
$begingroup$
the pdf is defined such that $f(x)dx$ gives the probability that $x$ will be measured in between $x$ and $x+dx$
$endgroup$
– phdmba7of12
Jan 28 at 19:19
1
$begingroup$
@phdmba7of12 thanks for your help... hmm it seems $t=e^x$? How do you get to the conclusion that $x=g^{-1}(t) = lnt$? Thanks again!
$endgroup$
– HJ_beginner
Jan 28 at 19:24
$begingroup$
$e^{ln z}=z$ just exponentiate both sides of the $x = ln t$ equation
$endgroup$
– phdmba7of12
Jan 28 at 19:25
$begingroup$
one way might be to use delta function as kernel for the transform: $$F(x)=int_0^x f(x)dx$$ and $$F(ln t)=int_0^{ln t}f(ln t)dleft(ln t right)$$
$endgroup$
– phdmba7of12
Jan 28 at 19:27
|
show 1 more comment
$begingroup$
$dt=tdx = e^xdx$
$endgroup$
– phdmba7of12
Jan 28 at 19:17
$begingroup$
the pdf is defined such that $f(x)dx$ gives the probability that $x$ will be measured in between $x$ and $x+dx$
$endgroup$
– phdmba7of12
Jan 28 at 19:19
1
$begingroup$
@phdmba7of12 thanks for your help... hmm it seems $t=e^x$? How do you get to the conclusion that $x=g^{-1}(t) = lnt$? Thanks again!
$endgroup$
– HJ_beginner
Jan 28 at 19:24
$begingroup$
$e^{ln z}=z$ just exponentiate both sides of the $x = ln t$ equation
$endgroup$
– phdmba7of12
Jan 28 at 19:25
$begingroup$
one way might be to use delta function as kernel for the transform: $$F(x)=int_0^x f(x)dx$$ and $$F(ln t)=int_0^{ln t}f(ln t)dleft(ln t right)$$
$endgroup$
– phdmba7of12
Jan 28 at 19:27
$begingroup$
$dt=tdx = e^xdx$
$endgroup$
– phdmba7of12
Jan 28 at 19:17
$begingroup$
$dt=tdx = e^xdx$
$endgroup$
– phdmba7of12
Jan 28 at 19:17
$begingroup$
the pdf is defined such that $f(x)dx$ gives the probability that $x$ will be measured in between $x$ and $x+dx$
$endgroup$
– phdmba7of12
Jan 28 at 19:19
$begingroup$
the pdf is defined such that $f(x)dx$ gives the probability that $x$ will be measured in between $x$ and $x+dx$
$endgroup$
– phdmba7of12
Jan 28 at 19:19
1
1
$begingroup$
@phdmba7of12 thanks for your help... hmm it seems $t=e^x$? How do you get to the conclusion that $x=g^{-1}(t) = lnt$? Thanks again!
$endgroup$
– HJ_beginner
Jan 28 at 19:24
$begingroup$
@phdmba7of12 thanks for your help... hmm it seems $t=e^x$? How do you get to the conclusion that $x=g^{-1}(t) = lnt$? Thanks again!
$endgroup$
– HJ_beginner
Jan 28 at 19:24
$begingroup$
$e^{ln z}=z$ just exponentiate both sides of the $x = ln t$ equation
$endgroup$
– phdmba7of12
Jan 28 at 19:25
$begingroup$
$e^{ln z}=z$ just exponentiate both sides of the $x = ln t$ equation
$endgroup$
– phdmba7of12
Jan 28 at 19:25
$begingroup$
one way might be to use delta function as kernel for the transform: $$F(x)=int_0^x f(x)dx$$ and $$F(ln t)=int_0^{ln t}f(ln t)dleft(ln t right)$$
$endgroup$
– phdmba7of12
Jan 28 at 19:27
$begingroup$
one way might be to use delta function as kernel for the transform: $$F(x)=int_0^x f(x)dx$$ and $$F(ln t)=int_0^{ln t}f(ln t)dleft(ln t right)$$
$endgroup$
– phdmba7of12
Jan 28 at 19:27
|
show 1 more comment
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$begingroup$
$dt=tdx = e^xdx$
$endgroup$
– phdmba7of12
Jan 28 at 19:17
$begingroup$
the pdf is defined such that $f(x)dx$ gives the probability that $x$ will be measured in between $x$ and $x+dx$
$endgroup$
– phdmba7of12
Jan 28 at 19:19
1
$begingroup$
@phdmba7of12 thanks for your help... hmm it seems $t=e^x$? How do you get to the conclusion that $x=g^{-1}(t) = lnt$? Thanks again!
$endgroup$
– HJ_beginner
Jan 28 at 19:24
$begingroup$
$e^{ln z}=z$ just exponentiate both sides of the $x = ln t$ equation
$endgroup$
– phdmba7of12
Jan 28 at 19:25
$begingroup$
one way might be to use delta function as kernel for the transform: $$F(x)=int_0^x f(x)dx$$ and $$F(ln t)=int_0^{ln t}f(ln t)dleft(ln t right)$$
$endgroup$
– phdmba7of12
Jan 28 at 19:27