Mixing
CDF of quotient of RVs?
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Let A and B be i.i.d exponential distributions with $lambda=1$. Let $C = A/B$. What is $P(C leq c)$
Let A and B be independent uniform [0,1] distributions. Let $C = A/B$. What is $P(C leq c)$
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 2 days ago
j doe
11
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j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
0
down vote
favorite
Let A and B be i.i.d exponential distributions with $lambda=1$. Let $C = A/B$. What is $P(C leq c)$
Let A and B be independent uniform [0,1] distributions. Let $C = A/B$. What is $P(C leq c)$
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 2 days ago
j doe
11
New contributor
j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
2 days ago
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let A and B be i.i.d exponential distributions with $lambda=1$. Let $C = A/B$. What is $P(C leq c)$
Let A and B be independent uniform [0,1] distributions. Let $C = A/B$. What is $P(C leq c)$
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 2 days ago
j doe
11
New contributor
j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let A and B be i.i.d exponential distributions with $lambda=1$. Let $C = A/B$. What is $P(C leq c)$
Let A and B be independent uniform [0,1] distributions. Let $C = A/B$. What is $P(C leq c)$
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 2 days ago
j doe
11
New contributor
j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
j doe
11
New contributor
j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 days ago
asked 2 days ago
j doe
11
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j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago
j doe
11
asked 2 days ago
j doe
11
11
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j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
j doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
2 days ago
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
2 days ago
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
2 days ago
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
2 days ago
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
2 days ago
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
2 days ago
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
2 days ago
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
2 days ago
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 2 days ago
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Graham Kemp
84k43378
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
2 days ago
Also for the second, would you be able to help with what the partitions are?
– j doe
2 days ago
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ 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
2 days ago
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
2 days ago
No, since the integral is not with respect to $c$.
– Graham Kemp
2 days ago
add a comment |
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 2 days ago
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Todor Markov
3365
answered 2 days ago
Todor Markov
3365
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Graham Kemp
84k43378
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
2 days ago
Also for the second, would you be able to help with what the partitions are?
– j doe
2 days ago
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ 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
2 days ago
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
2 days ago
No, since the integral is not with respect to $c$.
– Graham Kemp
2 days ago
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 2 days ago
Graham Kemp
84k43378
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
2 days ago
Also for the second, would you be able to help with what the partitions are?
– j doe
2 days ago
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ 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
2 days ago
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
2 days ago
No, since the integral is not with respect to $c$.
– Graham Kemp
2 days ago
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 2 days ago
Graham Kemp
84k43378
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 2 days ago
Graham Kemp
84k43378
answered 2 days ago
Graham Kemp
84k43378
answered 2 days ago
Graham Kemp
84k43378
answered 2 days ago
Graham Kemp
84k43378
84k43378
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
2 days ago
Also for the second, would you be able to help with what the partitions are?
– j doe
2 days ago
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ 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
2 days ago
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
2 days ago
No, since the integral is not with respect to $c$.
– Graham Kemp
2 days ago
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
2 days ago
Also for the second, would you be able to help with what the partitions are?
– j doe
2 days ago
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ 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
2 days ago
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
2 days ago
No, since the integral is not with respect to $c$.
– Graham Kemp
2 days ago
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
2 days ago
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
2 days ago
Also for the second, would you be able to help with what the partitions are?
– j doe
2 days ago
Also for the second, would you be able to help with what the partitions are?
– j doe
2 days ago
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ 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
2 days ago
$mathsf P(Cleqslant c)=mathsf E(e^{-A/c})$ 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
2 days ago
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
2 days ago
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
2 days ago
No, since the integral is not with respect to $c$.
– Graham Kemp
2 days ago
No, since the integral is not with respect to $c$.
– Graham Kemp
2 days ago
add a comment |
j doe is a new contributor. Be nice, and check out our Code of Conduct.
j doe is a new contributor. Be nice, and check out our Code of Conduct.
j doe is a new contributor. Be nice, and check out our Code of Conduct.
j doe is a new contributor. Be nice, and check out our Code of Conduct.
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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
2 days ago
Why would $mathsf P(Bgeqslant A/c)=mathsf E(e^{-1X/z})$ ? Where did the $X$ and $z$ come from?
– Graham Kemp
2 days ago