CDF of quotient of RVs?
up vote
0
down vote
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What would be density function for the ratio of X/Y, for both uniform and exponential distributions?
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-A/c}]$ but unsure what that is. I am stuck on the second.
probability probability-distributions
asked Nov 19 at 21:24
j doe
11
add a comment |
up vote
0
down vote
favorite
What would be density function for the ratio of X/Y, for both uniform and exponential distributions?
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-A/c}]$ but unsure what that is. I am stuck on the second.
probability probability-distributions
asked Nov 19 at 21:24
j doe
11
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 19 at 21:38
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
Nov 19 at 22:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What would be density function for the ratio of X/Y, for both uniform and exponential distributions?
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-A/c}]$ but unsure what that is. I am stuck on the second.
probability probability-distributions
asked Nov 19 at 21:24
j doe
11
What would be density function for the ratio of X/Y, for both uniform and exponential distributions?
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-A/c}]$ but unsure what that is. I am stuck on the second.
probability probability-distributions
probability probability-distributions
asked Nov 19 at 21:24
j doe
11
asked Nov 19 at 21:24
j doe
11
edited Nov 23 at 4:41
asked Nov 19 at 21:24
j doe
11
asked Nov 19 at 21:24
j doe
11
asked Nov 19 at 21:24
j doe
11
11
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 19 at 21:38
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
Nov 19 at 22:35
add a comment |
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 19 at 21:38
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
Nov 19 at 22:35
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 19 at 21:38
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 19 at 21:38
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
Nov 19 at 22:35
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
Nov 19 at 22:35
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Here's a hint:
- Can you solve the problem if $B$ was a constant, and not a random variable?
- How can you combine the above and the law of total probability to get your answer?
Image sourced from here
answered Nov 19 at 22:31
Todor Markov
65616
add a comment |
up vote
0
down vote
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-1X/z}]$ but unsure what that is.
When did random variable $X$ and constant $z$ enter into the discussion?
However, it is that: $$begin{align}mathsf P(Cleqslant c)&=mathsf P(Bgeqslant A/c)\&=mathsf E(mathsf P(Bgeqslant A/cmid A))\&=mathsf E(e^{-A/c})end{align}$$
The rest is just evaluating the expectation: $mathsf E(g(A))=int_Bbb R f_A(a)~g(a)~mathsf d a$ $$mathsf E(e^{-A/c})=int_0^infty e^{-a}~e^{-a/c}~mathsf d a$$
I am stuck on the second.
Second verse, much the same as the first.
Hint: it is a peicewise function partitioned on the magnitude of $c$. Consider the cases $0<c< 1$ and $1leqslant c$
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
So to be clear for the first one: 𝖤(e^−A/c) is the CDF and then would the pdf just be e^(-a)*e^(-a/c)?
– j doe
Nov 20 at 0:59
Also for the second, would you be able to help with what the partitions are?
– j doe
Nov 20 at 1:22
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ is the CDF of $C$. The pdf of $C$ is the derivative of that with respect to $c$. (Also, I clued you in on what the partitions should be. Consider those cases.)
– Graham Kemp
Nov 20 at 2:09
I am saying that if E(e^-A/c) is the CDF -- and that is the integral of some function f(a) -- can't I just say that f(a) (ie: the thing being integrated) is the PDF?
– j doe
Nov 20 at 2:34
No, since the integral is not with respect to $c$.
– Graham Kemp
Nov 20 at 2:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here's a hint:
- Can you solve the problem if $B$ was a constant, and not a random variable?
- How can you combine the above and the law of total probability to get your answer?
Image sourced from here
answered Nov 19 at 22:31
Todor Markov
65616
add a comment |
up vote
1
down vote
Here's a hint:
- Can you solve the problem if $B$ was a constant, and not a random variable?
- How can you combine the above and the law of total probability to get your answer?
Image sourced from here
answered Nov 19 at 22:31
Todor Markov
65616
add a comment |
up vote
1
down vote
up vote
1
down vote
Here's a hint:
- Can you solve the problem if $B$ was a constant, and not a random variable?
- How can you combine the above and the law of total probability to get your answer?
Image sourced from here
answered Nov 19 at 22:31
Todor Markov
65616
Here's a hint:
- Can you solve the problem if $B$ was a constant, and not a random variable?
- How can you combine the above and the law of total probability to get your answer?
Image sourced from here
answered Nov 19 at 22:31
Todor Markov
65616
answered Nov 19 at 22:31
Todor Markov
65616
answered Nov 19 at 22:31
Todor Markov
65616
answered Nov 19 at 22:31
Todor Markov
65616
65616
add a comment |
add a comment |
up vote
0
down vote
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-1X/z}]$ but unsure what that is.
When did random variable $X$ and constant $z$ enter into the discussion?
However, it is that: $$begin{align}mathsf P(Cleqslant c)&=mathsf P(Bgeqslant A/c)\&=mathsf E(mathsf P(Bgeqslant A/cmid A))\&=mathsf E(e^{-A/c})end{align}$$
The rest is just evaluating the expectation: $mathsf E(g(A))=int_Bbb R f_A(a)~g(a)~mathsf d a$ $$mathsf E(e^{-A/c})=int_0^infty e^{-a}~e^{-a/c}~mathsf d a$$
I am stuck on the second.
Second verse, much the same as the first.
Hint: it is a peicewise function partitioned on the magnitude of $c$. Consider the cases $0<c< 1$ and $1leqslant c$
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
So to be clear for the first one: 𝖤(e^−A/c) is the CDF and then would the pdf just be e^(-a)*e^(-a/c)?
– j doe
Nov 20 at 0:59
Also for the second, would you be able to help with what the partitions are?
– j doe
Nov 20 at 1:22
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ is the CDF of $C$. The pdf of $C$ is the derivative of that with respect to $c$. (Also, I clued you in on what the partitions should be. Consider those cases.)
– Graham Kemp
Nov 20 at 2:09
I am saying that if E(e^-A/c) is the CDF -- and that is the integral of some function f(a) -- can't I just say that f(a) (ie: the thing being integrated) is the PDF?
– j doe
Nov 20 at 2:34
No, since the integral is not with respect to $c$.
– Graham Kemp
Nov 20 at 2:37
add a comment |
up vote
0
down vote
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-1X/z}]$ but unsure what that is.
When did random variable $X$ and constant $z$ enter into the discussion?
However, it is that: $$begin{align}mathsf P(Cleqslant c)&=mathsf P(Bgeqslant A/c)\&=mathsf E(mathsf P(Bgeqslant A/cmid A))\&=mathsf E(e^{-A/c})end{align}$$
The rest is just evaluating the expectation: $mathsf E(g(A))=int_Bbb R f_A(a)~g(a)~mathsf d a$ $$mathsf E(e^{-A/c})=int_0^infty e^{-a}~e^{-a/c}~mathsf d a$$
I am stuck on the second.
Second verse, much the same as the first.
Hint: it is a peicewise function partitioned on the magnitude of $c$. Consider the cases $0<c< 1$ and $1leqslant c$
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
So to be clear for the first one: 𝖤(e^−A/c) is the CDF and then would the pdf just be e^(-a)*e^(-a/c)?
– j doe
Nov 20 at 0:59
Also for the second, would you be able to help with what the partitions are?
– j doe
Nov 20 at 1:22
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ is the CDF of $C$. The pdf of $C$ is the derivative of that with respect to $c$. (Also, I clued you in on what the partitions should be. Consider those cases.)
– Graham Kemp
Nov 20 at 2:09
I am saying that if E(e^-A/c) is the CDF -- and that is the integral of some function f(a) -- can't I just say that f(a) (ie: the thing being integrated) is the PDF?
– j doe
Nov 20 at 2:34
No, since the integral is not with respect to $c$.
– Graham Kemp
Nov 20 at 2:37
add a comment |
up vote
0
down vote
up vote
0
down vote
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-1X/z}]$ but unsure what that is.
When did random variable $X$ and constant $z$ enter into the discussion?
However, it is that: $$begin{align}mathsf P(Cleqslant c)&=mathsf P(Bgeqslant A/c)\&=mathsf E(mathsf P(Bgeqslant A/cmid A))\&=mathsf E(e^{-A/c})end{align}$$
The rest is just evaluating the expectation: $mathsf E(g(A))=int_Bbb R f_A(a)~g(a)~mathsf d a$ $$mathsf E(e^{-A/c})=int_0^infty e^{-a}~e^{-a/c}~mathsf d a$$
I am stuck on the second.
Second verse, much the same as the first.
Hint: it is a peicewise function partitioned on the magnitude of $c$. Consider the cases $0<c< 1$ and $1leqslant c$
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
For the first one, I think that $P(C⩽c)=P(B⩾A/c)=E[e^{-1X/z}]$ but unsure what that is.
When did random variable $X$ and constant $z$ enter into the discussion?
However, it is that: $$begin{align}mathsf P(Cleqslant c)&=mathsf P(Bgeqslant A/c)\&=mathsf E(mathsf P(Bgeqslant A/cmid A))\&=mathsf E(e^{-A/c})end{align}$$
The rest is just evaluating the expectation: $mathsf E(g(A))=int_Bbb R f_A(a)~g(a)~mathsf d a$ $$mathsf E(e^{-A/c})=int_0^infty e^{-a}~e^{-a/c}~mathsf d a$$
I am stuck on the second.
Second verse, much the same as the first.
Hint: it is a peicewise function partitioned on the magnitude of $c$. Consider the cases $0<c< 1$ and $1leqslant c$
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
answered Nov 19 at 23:20
Graham Kemp
84.6k43378
84.6k43378
So to be clear for the first one: 𝖤(e^−A/c) is the CDF and then would the pdf just be e^(-a)*e^(-a/c)?
– j doe
Nov 20 at 0:59
Also for the second, would you be able to help with what the partitions are?
– j doe
Nov 20 at 1:22
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ is the CDF of $C$. The pdf of $C$ is the derivative of that with respect to $c$. (Also, I clued you in on what the partitions should be. Consider those cases.)
– Graham Kemp
Nov 20 at 2:09
I am saying that if E(e^-A/c) is the CDF -- and that is the integral of some function f(a) -- can't I just say that f(a) (ie: the thing being integrated) is the PDF?
– j doe
Nov 20 at 2:34
No, since the integral is not with respect to $c$.
– Graham Kemp
Nov 20 at 2:37
add a comment |
So to be clear for the first one: 𝖤(e^−A/c) is the CDF and then would the pdf just be e^(-a)*e^(-a/c)?
– j doe
Nov 20 at 0:59
Also for the second, would you be able to help with what the partitions are?
– j doe
Nov 20 at 1:22
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ is the CDF of $C$. The pdf of $C$ is the derivative of that with respect to $c$. (Also, I clued you in on what the partitions should be. Consider those cases.)
– Graham Kemp
Nov 20 at 2:09
I am saying that if E(e^-A/c) is the CDF -- and that is the integral of some function f(a) -- can't I just say that f(a) (ie: the thing being integrated) is the PDF?
– j doe
Nov 20 at 2:34
No, since the integral is not with respect to $c$.
– Graham Kemp
Nov 20 at 2:37
So to be clear for the first one: 𝖤(e^−A/c) is the CDF and then would the pdf just be e^(-a)*e^(-a/c)?
– j doe
Nov 20 at 0:59
So to be clear for the first one: 𝖤(e^−A/c) is the CDF and then would the pdf just be e^(-a)*e^(-a/c)?
– j doe
Nov 20 at 0:59
Also for the second, would you be able to help with what the partitions are?
– j doe
Nov 20 at 1:22
Also for the second, would you be able to help with what the partitions are?
– j doe
Nov 20 at 1:22
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ is the CDF of $C$. The pdf of $C$ is the derivative of that with respect to $c$. (Also, I clued you in on what the partitions should be. Consider those cases.)
– Graham Kemp
Nov 20 at 2:09
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ is the CDF of $C$. The pdf of $C$ is the derivative of that with respect to $c$. (Also, I clued you in on what the partitions should be. Consider those cases.)
– Graham Kemp
Nov 20 at 2:09
I am saying that if E(e^-A/c) is the CDF -- and that is the integral of some function f(a) -- can't I just say that f(a) (ie: the thing being integrated) is the PDF?
– j doe
Nov 20 at 2:34
I am saying that if E(e^-A/c) is the CDF -- and that is the integral of some function f(a) -- can't I just say that f(a) (ie: the thing being integrated) is the PDF?
– j doe
Nov 20 at 2:34
No, since the integral is not with respect to $c$.
– Graham Kemp
Nov 20 at 2:37
No, since the integral is not with respect to $c$.
– Graham Kemp
Nov 20 at 2:37
add a comment |
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Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 19 at 21:38
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
Nov 19 at 22:35