How to calculate expected value and variance of a random variable [closed]
Let the random variable $Y$ have the following density:
$$f(y) = frac{1+beta y}2, -1 le y le 1, -1 le beta le 1$$
Find $E(Y)$ and $V(Y)$.
Can anyone help me with this problem?
probability statistics probability-distributions distribution-theory density-function
closed as off-topic by Davide Giraudo, heropup, KReiser, Cesareo, Chinnapparaj R Nov 22 '18 at 2:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, KReiser, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let the random variable $Y$ have the following density:
$$f(y) = frac{1+beta y}2, -1 le y le 1, -1 le beta le 1$$
Find $E(Y)$ and $V(Y)$.
Can anyone help me with this problem?
probability statistics probability-distributions distribution-theory density-function
closed as off-topic by Davide Giraudo, heropup, KReiser, Cesareo, Chinnapparaj R Nov 22 '18 at 2:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, KReiser, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
Hi, welcome to StackExchange! Generally we don't like to have questions in links. I've gone ahead and edited the question into the post.
– Monstrous Moonshiner
Nov 21 '18 at 22:58
Placing the question on hold seems awfully drastic. It is obvious the question is being put by someone who is just beginning a course in probability theory. The supplied answer is something that is usually given early in any probability course taken after elementary calculus.
– herb steinberg
Nov 22 '18 at 2:49
add a comment |
Let the random variable $Y$ have the following density:
$$f(y) = frac{1+beta y}2, -1 le y le 1, -1 le beta le 1$$
Find $E(Y)$ and $V(Y)$.
Can anyone help me with this problem?
probability statistics probability-distributions distribution-theory density-function
Let the random variable $Y$ have the following density:
$$f(y) = frac{1+beta y}2, -1 le y le 1, -1 le beta le 1$$
Find $E(Y)$ and $V(Y)$.
Can anyone help me with this problem?
probability statistics probability-distributions distribution-theory density-function
probability statistics probability-distributions distribution-theory density-function
edited Nov 21 '18 at 22:53


Monstrous Moonshiner
2,25011337
2,25011337
asked Nov 21 '18 at 21:56


Sadyraliev DiyarSadyraliev Diyar
9
9
closed as off-topic by Davide Giraudo, heropup, KReiser, Cesareo, Chinnapparaj R Nov 22 '18 at 2:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, KReiser, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Davide Giraudo, heropup, KReiser, Cesareo, Chinnapparaj R Nov 22 '18 at 2:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, KReiser, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
Hi, welcome to StackExchange! Generally we don't like to have questions in links. I've gone ahead and edited the question into the post.
– Monstrous Moonshiner
Nov 21 '18 at 22:58
Placing the question on hold seems awfully drastic. It is obvious the question is being put by someone who is just beginning a course in probability theory. The supplied answer is something that is usually given early in any probability course taken after elementary calculus.
– herb steinberg
Nov 22 '18 at 2:49
add a comment |
Hi, welcome to StackExchange! Generally we don't like to have questions in links. I've gone ahead and edited the question into the post.
– Monstrous Moonshiner
Nov 21 '18 at 22:58
Placing the question on hold seems awfully drastic. It is obvious the question is being put by someone who is just beginning a course in probability theory. The supplied answer is something that is usually given early in any probability course taken after elementary calculus.
– herb steinberg
Nov 22 '18 at 2:49
Hi, welcome to StackExchange! Generally we don't like to have questions in links. I've gone ahead and edited the question into the post.
– Monstrous Moonshiner
Nov 21 '18 at 22:58
Hi, welcome to StackExchange! Generally we don't like to have questions in links. I've gone ahead and edited the question into the post.
– Monstrous Moonshiner
Nov 21 '18 at 22:58
Placing the question on hold seems awfully drastic. It is obvious the question is being put by someone who is just beginning a course in probability theory. The supplied answer is something that is usually given early in any probability course taken after elementary calculus.
– herb steinberg
Nov 22 '18 at 2:49
Placing the question on hold seems awfully drastic. It is obvious the question is being put by someone who is just beginning a course in probability theory. The supplied answer is something that is usually given early in any probability course taken after elementary calculus.
– herb steinberg
Nov 22 '18 at 2:49
add a comment |
1 Answer
1
active
oldest
votes
$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.
add a comment |
$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.
add a comment |
$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.
$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.
answered Nov 21 '18 at 23:01
herb steinbergherb steinberg
2,4932310
2,4932310
add a comment |
add a comment |
Hi, welcome to StackExchange! Generally we don't like to have questions in links. I've gone ahead and edited the question into the post.
– Monstrous Moonshiner
Nov 21 '18 at 22:58
Placing the question on hold seems awfully drastic. It is obvious the question is being put by someone who is just beginning a course in probability theory. The supplied answer is something that is usually given early in any probability course taken after elementary calculus.
– herb steinberg
Nov 22 '18 at 2:49