Mixing
Probability theorem proof: adding a constant to each of the outcomes
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This is all I have so far... it sounds like a rather straight forward proof but I'm not really sure how to go about this. Also I don't understand how adding a constant to each outcome would be the same as just adding that constant at the end? If you add the constant to each outcome wouldn't that cause a greater result? Just trying some practice homework questions 
probability-theory
asked Jul 14 at 16:50
Lil
94431935
add a comment |Â
up vote
0
down vote
favorite
This is all I have so far... it sounds like a rather straight forward proof but I'm not really sure how to go about this. Also I don't understand how adding a constant to each outcome would be the same as just adding that constant at the end? If you add the constant to each outcome wouldn't that cause a greater result? Just trying some practice homework questions
probability-theory
asked Jul 14 at 16:50
Lil
94431935
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is all I have so far... it sounds like a rather straight forward proof but I'm not really sure how to go about this. Also I don't understand how adding a constant to each outcome would be the same as just adding that constant at the end? If you add the constant to each outcome wouldn't that cause a greater result? Just trying some practice homework questions
probability-theory
asked Jul 14 at 16:50
Lil
94431935
This is all I have so far... it sounds like a rather straight forward proof but I'm not really sure how to go about this. Also I don't understand how adding a constant to each outcome would be the same as just adding that constant at the end? If you add the constant to each outcome wouldn't that cause a greater result? Just trying some practice homework questions
probability-theory
asked Jul 14 at 16:50
Lil
94431935
asked Jul 14 at 16:50
Lil
94431935
asked Jul 14 at 16:50
Lil
94431935
asked Jul 14 at 16:50
Lil
94431935
94431935
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
The constant is added to the outcomes. You've added it to the product of outcome and probability. Your last line should say
$$
p_1(x_1+c)+p_2(x_2+c)+cdots+p_n(x_n+c);.
$$
Then you can collect all the terms with the constant and use $sum_ip_i=1$.
answered Jul 14 at 16:55
joriki
165k10180328
add a comment |Â
up vote
0
down vote
Let $X$ be a random variable which takes on the values $x_i$ with probability $p_i$ for $i=1,dotsc, n$. Then
$$
EX=sum_i=1^nx_ip_i
$$
and
$$
E(X+c)=sum_i=1^n(x_i+c)p_i=sum_i=1^nx_ip_i+csum_i=1^np_i=EX+c
$$
since the probabilities sum to one.
answered Jul 14 at 16:56
Foobaz John
18.1k41245
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The constant is added to the outcomes. You've added it to the product of outcome and probability. Your last line should say
$$
p_1(x_1+c)+p_2(x_2+c)+cdots+p_n(x_n+c);.
$$
Then you can collect all the terms with the constant and use $sum_ip_i=1$.
answered Jul 14 at 16:55
joriki
165k10180328
add a comment |Â
up vote
0
down vote
The constant is added to the outcomes. You've added it to the product of outcome and probability. Your last line should say
$$
p_1(x_1+c)+p_2(x_2+c)+cdots+p_n(x_n+c);.
$$
Then you can collect all the terms with the constant and use $sum_ip_i=1$.
answered Jul 14 at 16:55
joriki
165k10180328
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The constant is added to the outcomes. You've added it to the product of outcome and probability. Your last line should say
$$
p_1(x_1+c)+p_2(x_2+c)+cdots+p_n(x_n+c);.
$$
Then you can collect all the terms with the constant and use $sum_ip_i=1$.
answered Jul 14 at 16:55
joriki
165k10180328
The constant is added to the outcomes. You've added it to the product of outcome and probability. Your last line should say
$$
p_1(x_1+c)+p_2(x_2+c)+cdots+p_n(x_n+c);.
$$
Then you can collect all the terms with the constant and use $sum_ip_i=1$.
answered Jul 14 at 16:55
joriki
165k10180328
answered Jul 14 at 16:55
joriki
165k10180328
answered Jul 14 at 16:55
joriki
165k10180328
answered Jul 14 at 16:55
joriki
165k10180328
165k10180328
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $X$ be a random variable which takes on the values $x_i$ with probability $p_i$ for $i=1,dotsc, n$. Then
$$
EX=sum_i=1^nx_ip_i
$$
and
$$
E(X+c)=sum_i=1^n(x_i+c)p_i=sum_i=1^nx_ip_i+csum_i=1^np_i=EX+c
$$
since the probabilities sum to one.
answered Jul 14 at 16:56
Foobaz John
18.1k41245
add a comment |Â
up vote
0
down vote
Let $X$ be a random variable which takes on the values $x_i$ with probability $p_i$ for $i=1,dotsc, n$. Then
$$
EX=sum_i=1^nx_ip_i
$$
and
$$
E(X+c)=sum_i=1^n(x_i+c)p_i=sum_i=1^nx_ip_i+csum_i=1^np_i=EX+c
$$
since the probabilities sum to one.
answered Jul 14 at 16:56
Foobaz John
18.1k41245
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $X$ be a random variable which takes on the values $x_i$ with probability $p_i$ for $i=1,dotsc, n$. Then
$$
EX=sum_i=1^nx_ip_i
$$
and
$$
E(X+c)=sum_i=1^n(x_i+c)p_i=sum_i=1^nx_ip_i+csum_i=1^np_i=EX+c
$$
since the probabilities sum to one.
answered Jul 14 at 16:56
Foobaz John
18.1k41245
Let $X$ be a random variable which takes on the values $x_i$ with probability $p_i$ for $i=1,dotsc, n$. Then
$$
EX=sum_i=1^nx_ip_i
$$
and
$$
E(X+c)=sum_i=1^n(x_i+c)p_i=sum_i=1^nx_ip_i+csum_i=1^np_i=EX+c
$$
since the probabilities sum to one.
answered Jul 14 at 16:56
Foobaz John
18.1k41245
answered Jul 14 at 16:56
Foobaz John
18.1k41245
answered Jul 14 at 16:56
Foobaz John
18.1k41245
answered Jul 14 at 16:56
Foobaz John
18.1k41245
18.1k41245
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2851770%2fprobability-theorem-proof-adding-a-constant-to-each-of-the-outcomes%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password