For independent random variables α and β, is there a closed form expression for

E[α√α2+β2]

in terms of the expected values and variances of α and β? If not, is there a good lower bound on that expectation?

Update: I may as well mention that E[α]=1 and E[β]=0. I can control the variance on α and β, and I have in mind a setting where the variances of both α and β are pretty small relative to E[α]. Maybe both of their standard deviations are less than 0.3.

**Answer**

I thought of one lower bound, though I don’t think it’s very tight. I just pick an arbitrary value less than the mean of α and another arbitrary value around the mean of β2. Since the expectation is of a non-negative random variable, and because α and β are independent,

E[α√α2+β2]≥1√2P(α≥12)P(β2≤14).

By Chebyshev’s inequality,

P(α≥12)=P(α−1≥−12)≥P(|α−1|≤12)=1−P(|α−1|≥12)≥1−4var(α)

By Markov’s inequality,

P(β2≤14)=1−P(β2≥14)≥1−4E[β2]=1−4var(β)

Therefore,

E[α√α2+β2]≥1√2(1−4∗0.32)(1−4∗0.32)>0.28

Is a more standard/systematic way to do what I’m doing here, that gets a tighter bound?

**Attribution***Source : Link , Question Author : Jeff , Answer Author : Jeff*