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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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  • $\begingroup$ Sorry I should also add that the two random variables are independent. $\endgroup$ Commented Nov 7, 2015 at 19:57
  • $\begingroup$ Sorry to both you again, but if we have a constant c, can we say that E[f(cx)]=f(c)E[f(x)]? $\endgroup$ Commented Nov 7, 2015 at 20:19
  • $\begingroup$ I actually had a typo. The original expression shouldnt be E[B^ab] but rather E[(B^ab)^1/2]. Do I have to take a double integral now? After your comment I realized my mistake. $\endgroup$ Commented Nov 7, 2015 at 20:28
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    $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Commented Nov 7, 2015 at 21:30
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    $\begingroup$ Its not from either. Its a problem Im working on $\endgroup$ Commented Nov 7, 2015 at 21:31

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As indicated in my now erased comments \begin{align*}\mathbb{E}[B^ab]&=\mathbb{E}[B^a]/2\\&=\frac{1}{2}\int_0^1 B^a \text{d}a\\&=\frac{1}{2}\int_0^1 e^{\log(B)a} \text{d}a\\ &=\frac{1}{2}\frac{1}{\log(B)}\int_0^1 e^{\log(B)a} \log(B)\text{d}a\\&=\frac{1}{2\log(B)}[e^{\log(B)}-1]\\&=\frac{B-1}{2\log(B)}\end{align*} Now that the question has changed, $$\mathbb{E}[\sqrt{B^ab}]=\mathbb{E}[\sqrt{B}^a]\mathbb{E}[\sqrt{b}]=\frac{\sqrt{B}-1}{\log(B)}\,\frac{2}{3}$$

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  • $\begingroup$ As I have re-edited the question above, Its actually E[(B^a*b)/2)^0.5]. I am trying to re-work it. $\endgroup$ Commented Nov 7, 2015 at 22:03

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