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ChinG
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I had a question that I simiplified till the last step. However, I am not very comfortably withhave to calculate the expectations operator in a multiplicative/exponential setting.following: I know$$ E[a^{1/2}+b^{1/2}] $$ where $a=b=\frac{1}{2}\times10^{i}j$. We have that $a$$i$ is distributed uniformly ondistributed on say the $[0,1]$ interval and $b$$j$ is distrubtedalso uniformly distributed on the $[0,1].$ They$[0,1]$ interval. Both are both independent. HowI have canso far simplified this to: $$ E[2a^{1/2}]=2E[a^{^{1/2}}] $$ $E[a^{1/2}]$ is $E[(\frac{1}{2}\times10^{i}j)^{1/2})$. We know that $E[g(x,y)]=\int\int g(x,y)f(x,y)dxdy$. Substituting this for what we have, we obtain that $E[a^{1/2}]=\int\int(\frac{1}{2}10^{i}j)^{\frac{1}{2}}f(i,j)didj$. Given that $i$ and $j$ are independent, I simplifyobtain that the epxressionprevious $$ E[B^{a}b] $$expression simplifies to $\int\int\frac{1}{4}(10^{i}j)^{\frac{1}{2}}f(i)f(j)didj.$ Moreover, whereWe know that $B$ is a constant?$f(i)=f(j)=\frac{1}{2}.$ This then simplifies to $\frac{1}{16}\int\int10^{i}j\, didj$

I am stuck after this point and do not know the expectation operatorhow to proceed. Any help is linear, but the random variables are in exponential and multiplicative firm heremuch appreciated. Thanks a lot!

I had a question that I simiplified till the last step. However, I am not very comfortably with the expectations operator in a multiplicative/exponential setting. I know that $a$ is distributed uniformly on say $[0,1]$ and $b$ is distrubted uniformly on $[0,1].$ They are both independent. How can I simplify the epxression $$ E[B^{a}b] $$ where $B$ is a constant?

I know the expectation operator is linear, but the random variables are in exponential and multiplicative firm here. Thanks a lot!

I have to calculate the following: $$ E[a^{1/2}+b^{1/2}] $$ where $a=b=\frac{1}{2}\times10^{i}j$. We have that $i$ is uniformly distributed on say the $[0,1]$ interval and $j$ is also uniformly distributed on the $[0,1]$ interval. Both are independent. I have so far simplified this to: $$ E[2a^{1/2}]=2E[a^{^{1/2}}] $$ $E[a^{1/2}]$ is $E[(\frac{1}{2}\times10^{i}j)^{1/2})$. We know that $E[g(x,y)]=\int\int g(x,y)f(x,y)dxdy$. Substituting this for what we have, we obtain that $E[a^{1/2}]=\int\int(\frac{1}{2}10^{i}j)^{\frac{1}{2}}f(i,j)didj$. Given that $i$ and $j$ are independent, I obtain that the previous expression simplifies to $\int\int\frac{1}{4}(10^{i}j)^{\frac{1}{2}}f(i)f(j)didj.$ Moreover, We know that $f(i)=f(j)=\frac{1}{2}.$ This then simplifies to $\frac{1}{16}\int\int10^{i}j\, didj$

I am stuck after this point and do not know how to proceed. Any help is much appreciated.

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ChinG
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I had a question that I simiplified till the last step. However, I am not very comfortably with the expectations operator in a multiplicative/exponential setting. I know that $a$ is distributed uniformly on say $[0,1]$ and $b$ is distrubted uniformly on $[0,1].$ They are both independent. How can I simplify the epxression $$ E[B^{a}b] $$ where $B$ is a constant?

I know the expectation operator is linear, but the random variables are in exponential and multiplicative firm here. Thanks a lot!

I had a question that I simiplified till the last step. However, I am not very comfortably with the expectations operator in a multiplicative/exponential setting. I know that $a$ is distributed uniformly on say $[0,1]$ and $b$ is distrubted uniformly on $[0,1].$ How can I simplify the epxression $$ E[B^{a}b] $$ where $B$ is a constant?

I know the expectation operator is linear, but the random variables are in exponential and multiplicative firm here. Thanks a lot!

I had a question that I simiplified till the last step. However, I am not very comfortably with the expectations operator in a multiplicative/exponential setting. I know that $a$ is distributed uniformly on say $[0,1]$ and $b$ is distrubted uniformly on $[0,1].$ They are both independent. How can I simplify the epxression $$ E[B^{a}b] $$ where $B$ is a constant?

I know the expectation operator is linear, but the random variables are in exponential and multiplicative firm here. Thanks a lot!

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ChinG
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Expected Values in a Uniform Distribution

I had a question that I simiplified till the last step. However, I am not very comfortably with the expectations operator in a multiplicative/exponential setting. I know that $a$ is distributed uniformly on say $[0,1]$ and $b$ is distrubted uniformly on $[0,1].$ How can I simplify the epxression $$ E[B^{a}b] $$ where $B$ is a constant?

I know the expectation operator is linear, but the random variables are in exponential and multiplicative firm here. Thanks a lot!