How to calculate expected value and variance of a random variable [closed]

up vote
-1
down vote

favorite

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?

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

asked Nov 21 at 21:56

Sadyraliev Diyar

closed as off-topic by Davide Giraudo, heropup, KReiser, Cesareo, Chinnapparaj R Nov 22 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 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 at 2:49

add a comment |

up vote
-1
down vote

favorite

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?

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

asked Nov 21 at 21:56

Sadyraliev Diyar

closed as off-topic by Davide Giraudo, heropup, KReiser, Cesareo, Chinnapparaj R Nov 22 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 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 at 2:49

add a comment |

up vote
-1
down vote

favorite

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?

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

asked Nov 21 at 21:56

Sadyraliev Diyar

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

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

asked Nov 21 at 21:56

Sadyraliev Diyar

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

asked Nov 21 at 21:56

Sadyraliev Diyar

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

edited Nov 21 at 22:53

Monstrous Moonshiner

2,25511337

asked Nov 21 at 21:56

Sadyraliev Diyar

asked Nov 21 at 21:56

Sadyraliev Diyar

asked Nov 21 at 21:56

Sadyraliev Diyar

closed as off-topic by Davide Giraudo, heropup, KReiser, Cesareo, Chinnapparaj R Nov 22 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 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 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 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 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 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 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 at 2:49

add a comment |

1 Answer
1

active

oldest

votes

up vote
1
down vote

$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.

answered Nov 21 at 23:01

herb steinberg

2,4032310

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.

answered Nov 21 at 23:01

herb steinberg

2,4032310

add a comment |

up vote
1
down vote

$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.

answered Nov 21 at 23:01

herb steinberg

2,4032310

add a comment |

up vote
1
down vote

$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.

answered Nov 21 at 23:01

herb steinberg

2,4032310

$E(Y^k)=int_{-1}^1y^kf(y)dy$. $V(Y)=E(Y^2)-E(Y)^2$. I presume you can do the calculations.

answered Nov 21 at 23:01

herb steinberg

2,4032310

answered Nov 21 at 23:01

herb steinberg

2,4032310

answered Nov 21 at 23:01

herb steinberg

2,4032310

answered Nov 21 at 23:01

herb steinberg

2,4032310

add a comment |

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Krdytkyu