Compute $E[ \sqrt{Q} e^{-tQ} ]$ where $Q$ is non-central chi-square random variable.

Question

Let $Q$ be non-central chi-square random variable of degree $k$ and centrality parameter $\mu$.

How to compute the following expectation \begin{align} E\left[ \sqrt{Q} e^{-tQ} \right], t>0 \end{align}

I was thinking this can be computed by using the fact that $Q= \sqrt{ \sum_{i=1}^k Z_i^2 }$ where $Z_i \sim \mathcal{N}(\mu_i,1)$ and $\mu= \sqrt{\sum_{i=1}^k \mu_i^2}$, and $Z_i$'s are independent.

Using this observation we get that \begin{align} E\left[ \sqrt{\sum_{i=1}^k Z_i^2 } e^{-t\sum_{i=1}^k Z_i^2 } \right]= E\left[ \sqrt{\sum_{i=1}^k Z_i^2 } \prod_{i=1}^k e^{-tZ_i^2 } \right]. \end{align} However, it is not clear to me how such exressions can be manipulated or expanded.

I also tried direct integration but expression seem very long, so I stopped.

Answer 1

$$\mathbf E\left[ \sqrt{X} e^{-tX}\right] = e^{-\lambda/2} (1+2t)^{-(k+1)/2} \sqrt{2} \frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)} {}_1F_1 \left( \frac{k+1}{2}; \frac{k}{2}; \frac{\lambda}{2(1+2t)}\right) $$

---

We proceed to show this over a number of steps; first I outline the main claims, and then I will come back to prove the first of these (the rest are referenced below).

We consider a $\chi^2$ variable parameterised by $\lambda > 0$, $k > 0$ with PDF

$$ f(x) = \frac12 e^{-(x+\lambda)/2} \left( \frac{x}{\lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\lambda x}),$$ where $I$ denotes the modified Bessel function of the first kind.

Claim 1. (proof below) Let $X \sim \chi^2(\lambda,k)$ and $\tilde X \sim \chi^2(\tilde \lambda, k)$, where $\tilde \lambda = \lambda/(1+2t)$. Then

$$ \mathbf E\left[ \sqrt{X} e^{-tX}\right] = \exp\left(- \frac{t}{1+2t}\lambda\right) (1+2t)^{-(k+1)/2} \mathbf E \left[ \sqrt{ \tilde X}\right] $$

So in particular we have reduced the problem to calculating the expectation of the square root of a non-central chi-squared distribution. To do this we will want to relate the non-central $\chi^2$ distribution to the centralised one with $\lambda = 0$.

Claim 2. (Muirhead, 2005. Corollary 1.3.5) Let $P \sim \text{Poi}(\lambda/2)$ be Poisson distributed, then $$\chi^2(\lambda, k) = \chi^2(0,k+2 P).$$ Where the equality is in distribution.

In particular this claim gives us a way to calculate $\mathbf E[\sqrt{\tilde X}]$ by conditionining on $P$:

$$ \mathbf E[ \sqrt{\tilde X}] = e^{-\tilde \lambda/2} \sum_{p=0}^\infty \frac{(\tilde \lambda/2)^p}{p!} \mathbf E \left[\sqrt{X^{(k+2p)}}\right]$$ where I have denoted $X^{k} \sim \chi^2(0,k)$ for a centralised chi-squared distribution with $k$ degrees of freedom.

To tie things up we will require one further claim.

Claim 3. (Forbes, Evans, Hastings, Peacock 2011. Chapter 11.3) $$ \mathbf E \left[ \sqrt{X^{(k)}} \right] = \sqrt{2} \frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)}, $$

Piecing all of this together we have the main result:

Theorem $$\mathbf E\left[ \sqrt{X} e^{-tX}\right] = e^{-\lambda/2} (1+2t)^{-(k+1)/2} \sqrt{2} \sum_{p=0}^\infty \frac{\left(\frac{\lambda}{(2(1+2t)}\right)^p}{p!} \frac{\Gamma\left(p + \frac{k+1}{2}\right)}{\Gamma\left(p + \frac{k}{2}\right)} $$

The power series can be recognised as a confluent hypergeometric function, so we have the following corollary.

Corollary $$\mathbf E\left[ \sqrt{X} e^{-tX}\right] = e^{-\lambda/2} (1+2t)^{-(k+1)/2} \sqrt{2} \frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)} {}_1F_1 \left( \frac{k+1}{2}; \frac{k}{2}; \frac{\lambda}{2(1+2t)}\right) $$

---

Claims 2 and 3 are referenced above. Therefore I prove only Claim 1.

Following Raskolnikov's suggestion we turn to the PDF of the distribution, defined above. And note

$$ \begin{aligned} \mathbf E[ \sqrt{X} e^{-tX} ]& = \int x^{1/2} e^{-tx}f(x) dx \\ & = \int x^{1/2} \frac12 e^{-((1+2t)x+\lambda)/2} \left( \frac{x}{\lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\lambda x}) dx. \end{aligned} $$ Consider the change of variables $\tilde x = (1+2t)x$ and $\tilde \lambda = \lambda / (1+2t)$, then in particular note that the term in the exponential becomes

$$(1+2t)x + \lambda = \tilde x + (1+2t)\tilde \lambda,$$ whilst $$\left(\frac{x}{\lambda}\right)^{k/4 - 1/2} = \left(\frac1{(1+2t)^2} \frac{\tilde x}{\tilde \lambda}\right)^{k/4 -1/2} =(1+2t)^{1-k/2} \left(\frac{\tilde x}{\tilde \lambda}\right)^{k/4 -1/2} $$ and $$ \sqrt{\lambda x} = \sqrt{\tilde \lambda \tilde x}$$

Putting these together we have: $$ \begin{aligned} \mathbf E[ \sqrt{X} e^{-tX} ]& = (1+2t)^{1-k/2} \int x^{1/2} \frac12 \exp\left(-(\tilde x + (1+2t)\tilde \lambda)/2\right) \left( \frac{\tilde x}{\tilde \lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\tilde \lambda \tilde x}) dx \\ & = (1+2t)^{1-k/2} \exp(-t \tilde \lambda) \int x^{1/2} \frac12 \exp\left(-(\tilde x + \tilde \lambda)/2\right) \left( \frac{\tilde x}{\tilde \lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\tilde \lambda \tilde x}) dx \end{aligned} $$ We still have two terms which remain in terms of $x$, rather than $ \tilde x$:

$$ x^{1/2} = (1+2t)^{-1/2} \tilde x,$$ and to complete the change of variables formula

$$ dx = (1+2t)^{-1} d \tilde x.$$

Including these changes of variables we have

$$ \begin{aligned} \mathbf E[ \sqrt{X} e^{-tX} ]& = (1+2t)^{-(k+1)/2} \exp(-t \tilde \lambda) \int \tilde x^{1/2} \frac12 \exp\left(-(\tilde x + \tilde \lambda)/2\right) \left( \frac{\tilde x}{\tilde \lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\tilde \lambda \tilde x}) d \tilde x \\ & = (1+2t)^{-(k+1)/2} \exp(-t \tilde \lambda) \int \tilde x^{1/2} \tilde f(\tilde x) d \tilde x \end{aligned} $$ where $\tilde f$ denotes the PDF of the $\chi^2$ with parameters $\tilde \lambda$, and $k$. The result now follows on writing $$ \exp(-t \tilde \lambda) = \exp(-t\lambda/(1+2t))$$

---

Finally we check the above using a few different methods in R.

First we set some arbitrary choices of the parameters

t <- 1
lambda <- 2
k <- 2

In the first method, we sample $X$ directly as a $\chi^2(\lambda, k)$ distribution.

# Sample chi-squared.    
x <- rchisq(10e06, df = k, ncp = lambda)
mean( sqrt(x) * exp(-t *x) )
[1] 0.1436569

In the second we apply Claim 1, and sample the $\chi^2(\lambda/(1+2t,k)$ distribution

y <- rchisq(10e06, df = k, ncp = lambda/(1+2*t))
mean(sqrt(y)) * exp(- t * lambda/(1+2*t) ) * (1 + 2 *t)^{-(k + 1)/2}
[1] 0.1436606

In the third we apply Claim 2 and check that the Poisson weighting formula is correct.

df <- data_frame(p = rpois(10e06,lambda/(2*(1+2*t))) ) %>%
  mutate(
    z = rchisq(n(), df = k + 2 * p,ncp=0)
  )

mean(sqrt(df$z)) * exp(- t * lambda/(1 + 2 *t) ) * (1 + 2 *t)^{-(k + 1)/2}
[1] 0.1436255

Finally we compute the formula presented in the theorem statement.

library(CharFun) # Required for hypergeom1F1 function
exp(-lambda/2) *(1+2*t)^{-(k+1)/2} * sqrt(2) * gamma((k+1)/2) / gamma(k/2) * hypergeom1F1(lambda/(2*(1+2*t)), (k+1)/2, k/2)
[1] 0.1436613

Answer 2

Mathematica 11.3.0 produces

Mean[TransformedDistribution[Sqrt[x]*Exp[-t*x],x \[Distributed] 
 NoncentralChiSquareDistribution[k, \[Mu]],  Assumptions -> t > 0]]

$$\sqrt{2} e^{-\frac{\mu }{2}} (2 t+1)^{-\frac{k}{2}-\frac{1}{2}} \Gamma \left(\frac{k+1}{2}\right) \, _1\tilde{F}_1\left(\frac{k+1}{2};\frac{k}{2};\frac{\mu }{4 t+2}\right) $$

Edit. Thanks to @owen88 who paid my attention to the typo in my code:x*Exp[-t*x] instead of Sqrt[x]*Exp[-t*x]. Fixed.